Every rational number is
a natural number
an integer
a real number
a whole number
Between two rational numbers
there is no rational number
there is exactly one rational number
there are infinitely many rational numbers
there are only rational numbers and no irrational numbers
Decimal representation of a rational number cannot be
terminating
non-terminating
non-terminating repeating
non-terminating non-repeating
The product of any two irrational numbers is
always an irrational number
always a rational number
always an integer
sometimes rational, sometimes irrational
The decimal expansion of the number \(\sqrt{2}\) is
a finite decimal
\(1.41421\)
non-terminating recurring
non-terminating non-recurring
Which of the following is irrational?
\(\sqrt{\dfrac{4}{9}}\)
\(\dfrac{\sqrt{12}}{\sqrt{3}}\)
\(\sqrt{7}\)
\(\sqrt{81}\)
Which of the following is irrational?
\(0.14\)
\(0.14\overline{16}\)
\(0.\overline{1416}\)
\(0.4014001400014\ldots\)
A rational number between \(\sqrt{2}\) and \(\sqrt{3}\) is
\(\dfrac{\sqrt{2}+\sqrt{3}}{2}\)
\(\dfrac{\sqrt{2}\cdot\sqrt{3}}{2}\)
\(1.5\)
\(1.8\)
The value of \(1.999\ldots\) in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q\neq 0\), is
\(\dfrac{19}{10}\)
\(\dfrac{1999}{1000}\)
\(2\)
\(\dfrac{1}{9}\)
\(2\sqrt{3}+\sqrt{3}\) is equal to
\(2\sqrt{6}\)
\(6\)
\(3\sqrt{3}\)
\(4\sqrt{6}\)
\(\sqrt{10}\times\sqrt{15}\) is equal to
\(6\sqrt{5}\)
\(5\sqrt{6}\)
\(\sqrt{25}\)
\(10\sqrt{5}\)
The number obtained on rationalising the denominator of \(\dfrac{1}{\sqrt{7}-2}\) is
\(\dfrac{\sqrt{7}+2}{3}\)
\(\dfrac{\sqrt{7}-2}{3}\)
\(\dfrac{\sqrt{7}+2}{5}\)
\(\dfrac{\sqrt{7}+2}{45}\)
\(\dfrac{1}{\sqrt{9}-\sqrt{8}}\) is equal to
\(\dfrac{1}{2}(3-2\sqrt{2})\)
\(\dfrac{1}{3+2\sqrt{2}}\)
\(3-2\sqrt{2}\)
\(3+2\sqrt{2}\)
After rationalising the denominator of \(\dfrac{7}{3\sqrt{3}-2\sqrt{2}}\), we get the denominator as
\(13\)
\(19\)
\(5\)
\(35\)
The value of \(\dfrac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is equal to
\(\sqrt{2}\)
\(2\)
\(4\)
\(8\)
If \(\sqrt{2}=1.4142\), then \(\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) is equal to
\(2.4142\)
\(5.8282\)
\(0.4142\)
\(0.1718\)
\(\sqrt[4]{\sqrt[3]{2^{2}}}\) equals
\(2^{-\frac{1}{6}}\)
\(2^{-6}\)
\(2^{\frac{1}{6}}\)
\(2^{6}\)
The product \(\sqrt[3]{2}\cdot\sqrt[4]{2}\cdot\sqrt[12]{32}\) equals
\(\sqrt{2}\)
\(2\)
\(\sqrt[12]{2}\)
\(\sqrt[12]{32}\)
Value of \(\sqrt[4]{(81)^{-2}}\) is
\(\dfrac{1}{9}\)
\(\dfrac{1}{3}\)
\(9\)
\(\dfrac{1}{81}\)
Value of \((256)^{0.16}\times(256)^{0.09}\) is
\(4\)
\(16\)
\(64\)
\(256.25\)
Which of the following is equal to \(x\)?
\(x^{\frac{12}{7}}-x^{\frac{5}{7}}\)
\(\sqrt[12]{\left(x^{4}\right)^{\frac{1}{3}}}\)
\(\left(\sqrt{x^{3}}\right)^{\frac{2}{3}}\)
\(x^{\frac{12}{7}}\times x^{\frac{7}{12}}\)
Let \(x\) and \(y\) be rational and irrational numbers, respectively. Is \(x + y\) necessarily an irrational number? Give an example in support of your answer.
Yes. Let \(x = 21\), \(y = \sqrt{2}\) be a rational number.
Now \(x + y = 21 + \sqrt{2} = 21 + 1.4142\ldots = 22.4142\ldots\)
Which is non-terminating and non-recurring. Hence \(x + y\) is irrational.
Let \(x\) be rational and \(y\) be irrational. Is \(xy\) necessarily irrational? Justify your answer by an example.
No. \(0\times\sqrt{2}=0\) which is not irrational.
State whether the following statements are true or false? Justify your answer.
(i) \(\dfrac{\sqrt{2}}{3}\) is a rational number.
(ii) There are infinitely many integers between any two integers.
(iii) Number of rational numbers between 15 and 18 is finite.
(iv) There are numbers which cannot be written in the form \(\dfrac{p}{q}\), \(q\neq 0\), \(p,q\) both are integers.
(v) The square of an irrational number is always rational.
(vi) \(\dfrac{\sqrt{12}}{\sqrt{3}}\) is not a rational number as \(\sqrt{12}\) and \(\sqrt{3}\) are not integers.
(vii) \(\dfrac{\sqrt{15}}{\sqrt{3}}\) is written in the form \(\dfrac{p}{q}\), \(q\neq 0\) and so it is a rational number.
(i) False. Although \(\dfrac{\sqrt{2}}{3}\) is of the form \(\dfrac{p}{q}\) but here \(p\), i.e., \(\sqrt{2}\) is not an integer.
(ii) False. Between 2 and 3, there is no integer.
(iii) False, because between any two rational numbers we can find infinitely many rational numbers.
(iv) True. \(\dfrac{\sqrt{2}}{\sqrt{3}}\) is of the form \(\dfrac{p}{q}\) but \(p\) and \(q\) here are not integers.
(v) False, as \((\sqrt[4]{2})^2=\sqrt{2}\) which is not a rational number.
(vi) False, because \(\dfrac{\sqrt{12}}{\sqrt{3}}=\sqrt{4}=2\) which is a rational number.
(vii) False, because \(\dfrac{\sqrt{15}}{\sqrt{3}}=\sqrt{5}=\dfrac{\sqrt{5}}{1}\) which is \(p\), i.e., \(\sqrt{5}\) is not an integer.
Classify the following numbers as rational or irrational with justification:
(i) \(\sqrt{196}\)
(ii) \(3\sqrt{18}\)
(iii) \(\sqrt{\dfrac{9}{27}}\)
(iv) \(\dfrac{\sqrt{28}}{\sqrt{343}}\)
(v) \(-\sqrt{0.4}\)
(vi) \(\dfrac{\sqrt{12}}{\sqrt{75}}\)
(vii) \(0.5918\)
(viii) \((1+\sqrt{5})-(4+\sqrt{5})\)
(ix) \(10.124124\ldots\)
(x) \(1.010010001\ldots\)
(i) Rational, as \(\sqrt{196}=14\)
(ii) \(3\sqrt{18}=9\sqrt{2}\), which is the product of a rational and an irrational number and so an irrational number.
(iii) \(\sqrt{\dfrac{9}{27}}=\dfrac{1}{\sqrt{3}}\), which is the quotient of a rational and an irrational number and so an irrational number.
(iv) \(\dfrac{\sqrt{28}}{\sqrt{343}}=\dfrac{2}{7}\), which is a rational number.
(v) Irrational, \(-\sqrt{0.4}=-\dfrac{2}{\sqrt{10}}\), which is the quotient of a rational and an irrational.
(vi) \(\dfrac{\sqrt{12}}{\sqrt{75}}=\dfrac{2}{7}\), which is a rational number.
(vii) Rational, as decimal expansion is terminating.
(viii) \((1+\sqrt{5})-(4+\sqrt{5})=-3\), which is a rational number.
(ix) Rational, as decimal expansion is non-terminating recurring.
(x) Irrational, as decimal expansion is non-terminating non-recurring.
Find which of the variables \(x\), \(y\), \(z\) and \(u\) represent rational numbers and which represent irrational numbers:
(i) \(x^{2}=5\)
(ii) \(y^{2}=9\)
(iii) \(z^{2}=0.04\)
(iv) \(u^{2}=\dfrac{17}{4}\)
Rational numbers: (ii), (iii)
Irrational numbers: (i), (iv)
Find three rational numbers between:
(i) \(-1\) and \(-2\)
(ii) \(0.1\) and \(0.11\)
(iii) \(\dfrac{5}{7}\) and \(\dfrac{6}{7}\)
(iv) \(\dfrac{1}{4}\) and \(\dfrac{1}{5}\)
(i) \(-1.1\), \(-1.2\), \(-1.3\)
(ii) \(0.101\), \(0.102\), \(0.103\)
(iii) \(\dfrac{51}{70}\), \(\dfrac{52}{70}\), \(\dfrac{53}{70}\)
(iv) \(\dfrac{9}{40}\), \(\dfrac{17}{80}\), \(\dfrac{19}{80}\)
Insert a rational number and an irrational number between the following:
(i) \(2\) and \(3\)
(ii) \(0\) and \(0.1\)
(iii) \(\dfrac{1}{3}\) and \(\dfrac{1}{2}\)
(iv) \(-\dfrac{2}{5}\) and \(\dfrac{1}{2}\)
(v) \(0.15\) and \(0.16\)
(vi) \(\sqrt{2}\) and \(\sqrt{3}\)
(vii) \(2.357\) and \(3.121\)
(viii) \(0.0001\) and \(0.001\)
(ix) \(3.623623\) and \(0.484848\)
(x) \(6.375289\) and \(6.375738\)
(i) \(2.1\), \(2.040040004\ldots\)
(ii) \(0.03\), \(0.007000700007\ldots\)
(iii) \(\dfrac{5}{12}\), \(0.414114111\ldots\)
(iv) \(0\), \(0.151151115\ldots\)
(v) \(0.151\), \(0.151551555\ldots\)
(vi) \(1.5\), \(1.585585558\ldots\)
(vii) \(3\), \(3.101101110\ldots\)
(viii) \(0.00011\), \(0.00011331331333\ldots\)
(ix) \(1\), \(1.909009000\ldots\)
(x) \(6.3753\), \(6.375414114111\ldots\)
Represent the following numbers on the number line:
\(7\), \(7.2\), \(-\dfrac{3}{2}\), \(-\dfrac{12}{5}\)
Locate \(\sqrt{5}\), \(\sqrt{10}\) and \(\sqrt{17}\) on the number line.
Represent geometrically the following numbers on the number line:
(i) \(\sqrt{4.5}\)
(ii) \(\sqrt{5.6}\)
(iii) \(\sqrt{8.1}\)
(iv) \(\sqrt{2.3}\)
Express the following in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q\neq 0\):
(i) \(0.2\)
(ii) \(0.888\ldots\)
(iii) \(5.\overline{2}\)
(iv) \(0.\overline{001}\)
(v) \(0.2555\ldots\)
(vi) \(0.1\overline{34}\)
(vii) \(0.00323232\ldots\)
(viii) \(0.404040\ldots\)
(i) \(\dfrac{1}{5}\)
(ii) \(\dfrac{8}{9}\)
(iii) \(\dfrac{47}{9}\)
(iv) \(\dfrac{1}{999}\)
(v) \(\dfrac{23}{90}\)
(vi) \(\dfrac{133}{990}\)
(vii) \(\dfrac{8}{2475}\)
(viii) \(\dfrac{40}{99}\)
Show that \(0.142857142857\ldots = \dfrac{1}{7}\).
Simplify the following:
(i) \(\sqrt{45}-3\sqrt{20}+4\sqrt{5}\)
(ii) \(\dfrac{\sqrt{24}}{8}+\dfrac{\sqrt{54}}{9}\)
(iii) \(\sqrt[4]{12}\times\sqrt[7]{6}\)
(iv) \(4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}\)
(v) \(3\sqrt{3}+2\sqrt{27}+\dfrac{7}{\sqrt{3}}\)
(vi) \((\sqrt{3}-\sqrt{2})^{2}\)
(vii) \(\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}\)
(viii) \(\dfrac{3}{\sqrt{8}}+\dfrac{1}{\sqrt{2}}\)
(ix) \(\dfrac{2\sqrt{3}}{3}-\dfrac{\sqrt{3}}{6}\)
(i) \(\sqrt{5}\)
(ii) \(\dfrac{7\sqrt{6}}{12}\)
(iii) \(168\sqrt{2}\)
(iv) \(\dfrac{8}{3}\)
(v) \(\dfrac{34\sqrt{3}}{3}\)
(vi) \(5-2\sqrt{6}\)
(vii) \(0\)
(viii) \(\dfrac{5}{4}\sqrt{2}\)
(ix) \(\dfrac{\sqrt{3}}{2}\)
Rationalise the denominator of the following:
(i) \(\dfrac{2}{3\sqrt{3}}\)
(ii) \(\dfrac{\sqrt{40}}{\sqrt{3}}\)
(iii) \(\dfrac{3+\sqrt{2}}{4\sqrt{2}}\)
(iv) \(\dfrac{16}{\sqrt{41}-5}\)
(v) \(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\)
(vi) \(\dfrac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
(vii) \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
(viii) \(\dfrac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
(ix) \(\dfrac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\)
(i) \(\dfrac{2\sqrt{3}}{9}\)
(ii) \(\dfrac{2\sqrt{30}}{3}\)
(iii) \(\dfrac{2+3\sqrt{2}}{8}\)
(iv) \(\sqrt{41}+5\)
(v) \(7+4\sqrt{3}\)
(vi) \(3\sqrt{2}-2\sqrt{3}\)
(vii) \(5+2\sqrt{6}\)
(viii) \(9+2\sqrt{15}\)
(ix) \(\dfrac{9+4\sqrt{6}}{15}\)
Find the values of \(a\) and \(b\) in each of the following:
(i) \(\dfrac{5+2\sqrt{3}}{7+4\sqrt{3}}=a-6\sqrt{3}\)
(ii) \(\dfrac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\dfrac{19}{11}\)
(iii) \(\dfrac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-b\sqrt{6}\)
(iv) \(\dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}=a+\dfrac{7}{11}\sqrt{5}\,b\)
(i) \(a=11\)
(ii) \(a=\dfrac{9}{11}\)
(iii) \(b=-\dfrac{5}{6}\)
(iv) \(a=0\), \(b=1\)
If \(a=2+\sqrt{3}\), then find the value of \(a-\dfrac{1}{a}\).
\(2\sqrt{3}\)
Rationalise the denominator in each of the following and hence evaluate by taking \(\sqrt{2}=1.414\), \(\sqrt{3}=1.732\) and \(\sqrt{5}=2.236\), up to three places of decimal:
(i) \(\dfrac{4}{\sqrt{3}}\)
(ii) \(\dfrac{6}{\sqrt{6}}\)
(iii) \(\dfrac{\sqrt{10}-\sqrt{5}}{2}\)
(iv) \(\dfrac{\sqrt{2}}{2+\sqrt{2}}\)
(v) \(\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
(i) \(2.309\)
(ii) \(2.449\)
(iii) \(0.463\)
(iv) \(0.414\)
(v) \(0.318\)
Simplify:
(i) \(\left(1^{3}+2^{3}+3^{3}\right)^{1/2}\)
(ii) \(\left(\dfrac{3}{5}\right)^{4}\left(\dfrac{8}{5}\right)^{-12}\left(\dfrac{32}{5}\right)^{6}\)
(iii) \(\left(\dfrac{1}{27}\right)^{-2/3}\)
(iv) \(\left(\left(625\right)^{-1/2}\right)^{-1/4}\)\(^{2}\)
(v) \(\dfrac{9^{1/3}\times 27^{1/2}}{3^{1/2}\times 3^{2}}\)
(vi) \(64^{-1/3}\left(64^{1/3}-64^{2/3}\right)\)
(vii) \(\dfrac{8^{1/3}\times 16^{1/3}}{32^{-1/3}}\)
(i) \(6\)
(ii) \(\dfrac{2025}{64}\)
(iii) \(9\)
(iv) \(5\)
(v) \(3^{-1/3}\)
(vi) \(-3\)
(vii) \(16\)
Express \(0.6 + 0.\overline{7} + 0.4\overline{7}\) in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\).
\(0.6 = \dfrac{6}{10} = \dfrac{3}{5}\).
Let \(0.\overline{7} = x\). Then \(10x = 7.\overline{7}\).
Subtracting, \(10x - x = 7.\overline{7} - 0.\overline{7}\Rightarrow 9x = 7\Rightarrow x = \dfrac{7}{9}\).
Let \(0.4\overline{7} = y\). Then \(y = 0.4 + 0.0\overline{7}\).
\(0.4 = \dfrac{4}{10} = \dfrac{2}{5}\) and \(0.0\overline{7} = \dfrac{1}{10}\cdot 0.\overline{7} = \dfrac{1}{10}\cdot\dfrac{7}{9} = \dfrac{7}{90}\).
So \(y = \dfrac{2}{5} + \dfrac{7}{90} = \dfrac{36}{90} + \dfrac{7}{90} = \dfrac{43}{90}\).
Now, \(0.6 + 0.\overline{7} + 0.4\overline{7} = \dfrac{3}{5} + \dfrac{7}{9} + \dfrac{43}{90}\).
Taking LCM \(= 90\): \(\dfrac{3}{5} = \dfrac{54}{90}\), \(\dfrac{7}{9} = \dfrac{70}{90}\).
Sum \(= \dfrac{54}{90} + \dfrac{70}{90} + \dfrac{43}{90} = \dfrac{167}{90}\).
Simplify:
\(\dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\dfrac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}\).
Rationalise each term using the conjugate of the denominator.
\(\dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}} = \dfrac{7\sqrt{3}(\sqrt{10}-\sqrt{3})}{(\sqrt{10}+\sqrt{3})(\sqrt{10}-\sqrt{3})} = \dfrac{7\sqrt{3}(\sqrt{10}-\sqrt{3})}{10-3}\).
So \(\dfrac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}} = \sqrt{3}(\sqrt{10}-\sqrt{3}) = \sqrt{30}-3\).
\(\dfrac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}} = \dfrac{2\sqrt{5}(\sqrt{6}-\sqrt{5})}{6-5} = 2\sqrt{5}(\sqrt{6}-\sqrt{5}) = 2\sqrt{30}-10\).
\(\dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} = \dfrac{3\sqrt{2}(\sqrt{15}-3\sqrt{2})}{15-(3\sqrt{2})^2} = \dfrac{3\sqrt{2}(\sqrt{15}-3\sqrt{2})}{15-18} = \dfrac{3\sqrt{2}(\sqrt{15}-3\sqrt{2})}{-3}\).
So \(\dfrac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}} = -\sqrt{2}(\sqrt{15}-3\sqrt{2}) = -(\sqrt{30}-6) = -\sqrt{30}+6\).
Now the given expression becomes:
\((\sqrt{30}-3) - (2\sqrt{30}-10) - (-\sqrt{30}+6)\).
= \(\sqrt{30}-3-2\sqrt{30}+10+\sqrt{30}-6\).
= \((\sqrt{30}-2\sqrt{30}+\sqrt{30}) + ( -3+10-6)\) = \(0 + 1\) = \(1\).
If \(\sqrt{2}=1.414\) and \(\sqrt{3}=1.732\), find the value of
\(\dfrac{4}{3\sqrt{3}-2\sqrt{2}}+\dfrac{3}{3\sqrt{3}+2\sqrt{2}}\).
Rationalise each fraction using the conjugate of the denominator.
\(\dfrac{4}{3\sqrt{3}-2\sqrt{2}} = \dfrac{4(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2} = \dfrac{4(3\sqrt{3}+2\sqrt{2})}{27-8} = \dfrac{4(3\sqrt{3}+2\sqrt{2})}{19}\).
\(\dfrac{3}{3\sqrt{3}+2\sqrt{2}} = \dfrac{3(3\sqrt{3}-2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2} = \dfrac{3(3\sqrt{3}-2\sqrt{2})}{27-8} = \dfrac{3(3\sqrt{3}-2\sqrt{2})}{19}\).
Add them:
\(\dfrac{4(3\sqrt{3}+2\sqrt{2})}{19}+\dfrac{3(3\sqrt{3}-2\sqrt{2})}{19} = \dfrac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{19}\).
= \(\dfrac{21\sqrt{3}+2\sqrt{2}}{19}\).
Now substitute \(\sqrt{3}=1.732\), \(\sqrt{2}=1.414\):
Numerator \(= 21(1.732) + 2(1.414) = 36.372 + 2.828 = 39.200\).
Value \(= \dfrac{39.200}{19} = 2.063157\ldots \approx 2.063\).
If \(a=\dfrac{3+\sqrt{5}}{2}\), find the value of \(a^2+\dfrac{1}{a^2}\).
First find \(\dfrac{1}{a}\).
\(a=\dfrac{3+\sqrt{5}}{2}\Rightarrow \dfrac{1}{a}=\dfrac{2}{3+\sqrt{5}}\).
Rationalise:
\(\dfrac{2}{3+\sqrt{5}}\cdot\dfrac{3-\sqrt{5}}{3-\sqrt{5}} = \dfrac{2(3-\sqrt{5})}{9-5} = \dfrac{2(3-\sqrt{5})}{4} = \dfrac{3-\sqrt{5}}{2}\).
So \(a+\dfrac{1}{a} = \dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2} = \dfrac{6}{2}=3\).
Square both sides:
\((a+\dfrac{1}{a})^2 = a^2 + 2 + \dfrac{1}{a^2}\).
\(3^2 = a^2 + 2 + \dfrac{1}{a^2}\Rightarrow 9 = a^2 + \dfrac{1}{a^2} + 2\).
Therefore, \(a^2 + \dfrac{1}{a^2} = 7\).
If \(x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) and \(y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\), find the value of \(x^2+y^2\).
Note that \(xy=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=1\). Hence \(y=\dfrac{1}{x}\).
Now compute \(x+y\):
\(x+y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\).
Common denominator \((\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = 3-2 = 1\).
So \(x+y = (\sqrt{3}+\sqrt{2})^2 + (\sqrt{3}-\sqrt{2})^2\).
\((\sqrt{3}+\sqrt{2})^2 = 3+2+2\sqrt{6}=5+2\sqrt{6}\).
\((\sqrt{3}-\sqrt{2})^2 = 3+2-2\sqrt{6}=5-2\sqrt{6}\).
Therefore \(x+y = (5+2\sqrt{6})+(5-2\sqrt{6})=10\).
Since \(y=\dfrac{1}{x}\), we have \(x+\dfrac{1}{x}=10\).
Square both sides:
\((x+\dfrac{1}{x})^2 = x^2 + 2 + \dfrac{1}{x^2} = x^2 + y^2 + 2\).
\(10^2 = x^2 + y^2 + 2\Rightarrow 100 = x^2 + y^2 + 2\).
Hence \(x^2 + y^2 = 98\).
Simplify: \((256)^{-(4^{-3/2})}\).
Evaluate \(4^{-3/2}\):
\(4^{-3/2} = \dfrac{1}{4^{3/2}} = \dfrac{1}{(\sqrt{4})^3} = \dfrac{1}{2^3} = \dfrac{1}{8}\).
So the exponent becomes \(-(4^{-3/2}) = -\dfrac{1}{8}\).
Thus \((256)^{-(4^{-3/2})} = 256^{-1/8}\).
Write \(256\) as a power of 2: \(256 = 2^8\).
Then \(256^{-1/8} = (2^8)^{-1/8} = 2^{-1} = \dfrac{1}{2}\).
Find the value of
\(\dfrac{4}{(216)^{-2/3}}+\dfrac{1}{(256)^{-3/4}}+\dfrac{2}{(243)^{-1/5}}\).
Use \(\dfrac{1}{a^{-m}} = a^{m}\).
\(\dfrac{4}{(216)^{-2/3}} = 4\cdot 216^{2/3}\).
\(216 = 6^3\Rightarrow 216^{2/3} = (6^3)^{2/3} = 6^2 = 36\).
So first term \(= 4\cdot 36 = 144\).
\(\dfrac{1}{(256)^{-3/4}} = 256^{3/4}\).
\(256 = 2^8\Rightarrow 256^{3/4} = (2^8)^{3/4} = 2^{6} = 64\).
\(\dfrac{2}{(243)^{-1/5}} = 2\cdot 243^{1/5}\).
\(243 = 3^5\Rightarrow 243^{1/5} = 3\). So third term \(= 2\cdot 3 = 6\).
Total \(= 144 + 64 + 6 = 214\).