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 \(= 144 + 64 + 6 = 214\).