(4000000)³\(\div\)(0.00002)⁴ \(=(4.0 \times 10^6)^3\div(2.0 \times 10^{-5})^4\)
\(=(4.0)^3\times (10^6)^3 \div (2.0)^4\times (10^{-5})^4\)
\(={64.0\times 10^18\over 16.0 \times 10^{-20}}\)
=4 x 10¹⁸ x 10⁺²⁰
=4.0 x 10³⁸

மொத்த நிறை = 5.97 x 10²⁴கி.கி + 0.073 x 10²⁴கி.கி.
= (5.97 + 0.073) x 10²⁴ கி.கி.
= 6.043 x 10²⁴ கி.கி.

\(\begin{aligned} x & =\sqrt{5}+2 \end{aligned}\)
\(\begin{aligned} \frac{1}{x} & =\frac{1}{\sqrt{5}+2}=\frac{1}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{\sqrt{5}^2-2^2}=\frac{\sqrt{5}-2}{5-4}=\frac{\sqrt{5}-2}{1} \\ \end{aligned}\)
\(\begin{aligned} x+\frac{1}{x} & =\sqrt{5}+\not 2+\sqrt{5}-\not 2=2 \sqrt{5} \end{aligned}\)
\(\begin{aligned} \left(x+\frac{1}{x}\right)^2 & =(2 \sqrt{5})^2=4 \times 5=20 \end{aligned}\)
\(\begin{aligned} x^2+\frac{1}{x^2} & =\left(x+\frac{1}{x}\right)^2-2 \not x \times \frac{1}{\not x}=20-2=18 \end{aligned}\)

\({\sqrt[9]{8}\over \sqrt[6]{6}}={8^{1\over9}\over 6^{1\over6}}\) (6,9 இன் மீ.பொ.ம18என்பதைக் குறித்துக் கொள்க.)
\(= {8^{2\over 18}\over 6^{3\over18}}\)(எப்படி ?) \(=({8^2\over 6^3})^{1\over18}=({8\times8\over 6\times6\times6})^{1\over18}\)
\(=({8\over27})^{1\over18}=[({2\over3})^3]^{1\over18}=({2\over3})^{1\over6}=\sqrt[6]{2\over3}\)

\(\sqrt[3]{2},\sqrt[2]{4},\sqrt[4]{3}\) ஆகியவற்றின் வரிசைகள் 3,2.4
3,2,4 இன் மீ.பொ.ம =12.
\(\sqrt[3]{2}=(2^{1\over3})=({2^{4\over12}})=\sqrt[12]{2^4}=\sqrt[12]{16};\)
\(\sqrt[2]{4}=(4^{1\over2})=({4^{6\over12}})=\sqrt[12]{4^6} =\sqrt[12]{4096}\)
\(\sqrt[4]{3}=(3^{1\over4})=(3^{3\over12})=\sqrt[12]{3^3}=\sqrt[12]{27}\)
\(\sqrt[3]{2},\sqrt[4]{3},\sqrt[2]{4}\) ஆனது \(\sqrt[12]{16}<\sqrt[12]{27}<\sqrt[12]{4096}\)
ஏறுவரிசை \(\sqrt[3]{2},\sqrt[4]{3},\sqrt[2]{4}\)

(1) (i) \(\sqrt{8}=\sqrt{4\times2}=\sqrt[2]{2}\)
(ii) \(\sqrt[3]{192}=\sqrt[3]{4\times 4\times 4\times 3}=4\sqrt[3]{3}\)
(2) \(\sqrt[3]{7}=\sqrt[12]{7^4}=\sqrt[12]{2401}\)
\(\sqrt[4]{5}=5^{1\over4}=5^{3\over12}=\sqrt[12]{5^3}=\sqrt[12]{125}\)
\(\sqrt[12]{2401}>\sqrt[12]{125}\)
ஆகவே,\(\sqrt[3]{7}>\sqrt[4]{5}\).

(i)\(\begin{aligned} \sqrt{5} & =5^{\frac{1}{2}} \end{aligned}\)
(ii) \(\begin{aligned} \sqrt[2]{7} & =7^{\frac{1}{2}} \end{aligned}\)
(iii) \(\begin{aligned} (\sqrt[3]{49})^5 & =49^{\frac{5}{3}}=\left(7^2\right)^{\frac{5}{3}}=7^{\frac{10}{3}} \\ \end{aligned}\)
(iv)\(\begin{aligned} \left(\frac{1}{\sqrt[3]{100}}\right)^7 & =\left(\frac{1}{100^{\frac{1}{3}}}\right)^7=\left(\frac{1^{\frac{1}{3}}}{100^{\frac{1}{3}}}\right)^7=\left(\frac{1}{100}\right)^{\frac{7}{3}}=\left(\frac{1}{10^2}\right)^{\frac{7}{3}}=\left(10^{-2}\right)^{\frac{7}{3}}=10^{\frac{-14}{3}} \end{aligned}\)

\(64^{-2\over3}={1\over 64^{2\over3}}={1\over (\sqrt[3]{64})^2}={1\over4^2}\) (எப்படி?) =\({1\over16}\)

\(\sqrt{2}=2^{1/2}\)

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

(i) 8 = 2 × 2 × 2; எனவே  8 = 2³
(ii) 32 = 2 x 2 x 2 x 2 x 2 = 2⁵
(iii) \({1\over4}\)
\({1\over4}={1\over 2\times 2}={1\over 2^2}=2^{-2}\)
(iv) \(\sqrt{2}=2^{1/2}\)
(v) \(\sqrt{8}\)
\(\sqrt{8}=\sqrt{2}\times\sqrt2 \times \sqrt2=({2}^{1\over2})^3\). இதனை \(2^{3\over2}\)எனவும் எழுதலாம்.

\({\sqrt{5}\over \sqrt{6}+2}-{\sqrt{5}\over \sqrt{6}-2}\) \(=\frac{\sqrt{5}}{\sqrt{6}+2}-\frac{\sqrt{6}-2}{\sqrt{6}-2} \times \frac{\sqrt{5}}{\sqrt{6}-2} \times \frac{\sqrt{6}+2}{\sqrt{6}+2} \)
\(=\frac{\sqrt{30}-2 \sqrt{5}}{6-4}-\frac{\sqrt{30}+2 \sqrt{5}}{6-4} \)
\(=\frac{\sqrt{30}-2 \sqrt{5}-\sqrt{30}+2 \sqrt{5}}{2}=\frac{-4 \sqrt{5}}{2} \)
\(=-2 \sqrt{5} \)

\({5\sqrt{3}+\sqrt{2}\over \sqrt{3}+\sqrt{2}}\) \(=\frac{5 \sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \)
\(=\frac{(5 \sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}{3-2} \)
\(=\frac{5 \times 3-5 \sqrt{6}+\sqrt{6}-2}{1} \)
\(=\frac{15-4 \sqrt{6}-2}{1}=13-4 \sqrt{6} \)

(i) \(\begin{aligned} \sqrt{40}-\sqrt{20} & =\sqrt{2^2 \times 10}-\sqrt{2^2 \times 5} \end{aligned}\)
\(\begin{aligned} =(2 \times \sqrt{10})-(2 \times \sqrt{5}) \end{aligned}\)
\(\begin{aligned} =(2 \times 3.162)-(2 \times 2.236) \end{aligned}\)
= 6.324 - 4.472
= 1.852
(ii) \(\begin{aligned} \sqrt{300}+\sqrt{90}-\sqrt{8} & =\sqrt{3 \times 100}+\sqrt{9 \times 10}+\sqrt{4 \times 2}=10 \sqrt{3}+3 \sqrt{10}-2 \sqrt{2} \\ \end{aligned}\)
\( \begin{aligned} =10 \times 1.732+3 \times 3.162-2 \times 1.414 \end{aligned}\)
= 17.32 + 9.486 - 2.828 = 23.978

(i) \(\sqrt{3}=3^{1\over2}\)
\(=3^{6\over12}\)
\(=\sqrt[12]{3^6}\)
\(=\sqrt[12]{729}\)
(ii) \(\sqrt[4]{3}=3^{1\over4}\)
\(=3^{3\over12}\)
\(=\sqrt[12]{3^3}\)
\(=\sqrt[12]{27}\)
(iii) \(\sqrt[3]{3}=3^{1\over3}\)
\(=3^{4\over12}\)
\(=\sqrt[12]{3^4}\)
\(=\sqrt[12]{81}\)

(i) இயலும் 
எ.கா :\(4^ 3\sqrt { 21 } + (-3 ^3\sqrt {21}) = 3\sqrt {21}\) ஒரு முழுமையான முறுடு
(i) இயலும். 
எ.கா : \(7^ 4\sqrt { 25 } - 6 ^4\sqrt {25} = 4\sqrt {25}(7 - 6) = 4\sqrt {25}\) ஒரு முழுமையான முறுடு
(iii) இயலும் 
எ.கா : \(3\sqrt { 5 } \times 3 \sqrt { 4 } =3\sqrt { 20 } \) ஒரு முழுமையான முறுடு
(iv) இயலும் 
எ.கா :  \(\sqrt { \frac { 10 }{ 2 } } =\frac {\sqrt { 2 \times 5 }}{\sqrt {2}} =\frac {\sqrt {2} \times \sqrt {5}}{\sqrt{2}} = 5 \), ஒரு முழுமையான முறுடு

(i) \(\sqrt{3} \times \sqrt{5} \times \sqrt{2}=\sqrt{3 \times 5 \times 2}=\sqrt{30}\)
(ii) \(\sqrt{35} \div \sqrt{7}=\sqrt{\frac{35}{7}}=\sqrt{5}\)
(iii) \(\sqrt[3]{27} \times \sqrt[3]{8} \times \sqrt[3]{125}=\sqrt[3]{27 \times 8 \times 125}=\sqrt[3]{3^3 \times 2^3 \times 5^3}=3 \times 2 \times 5=30\)
(iv) \((7 \sqrt{a}-5 \sqrt{b})(7 \sqrt{a}+5 \sqrt{b})=(7 \sqrt{a})^2-(5 \sqrt{b})^2=49 a-25 b\)
(v)\(\begin{aligned} {\left[\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}\right] \div \sqrt{\frac{16}{81}} } & =\left[\sqrt{\frac{15^2}{27^2}}-\sqrt{\frac{5^2}{12^2}}\right] \times \sqrt{\frac{9^2}{4^2}} \end{aligned}\)
\(\begin{aligned} =\left(\frac{15}{27}-\frac{5}{12}\right) \times \frac{9}{4}=\left(\frac{5}{9}-\frac{5}{12}\right) \times \frac{9}{4} \end{aligned}\)

l = 5 x 10⁵
b = 4 x 10⁴ 
பரப்பளவு = l x b 
5 x 10⁵ x 4 x 10⁴ 
= 20 x 10⁵⁺⁴
= 20 x 10⁹ = 2.0 x 10¹ x 10⁹ = 2 x 10¹⁰மீ²

\((2 \sqrt{5}-\sqrt{2})^2=4 \times 5-2 \times 2 \sqrt{5} \times \sqrt{2}+2 \)
\(20-4 \sqrt{10}+2=22-4 \sqrt{10} \)

\((\because 4 \sqrt{7} \times 2 \sqrt{3}=4 \times 2 \sqrt{7 \times 3}=8 \sqrt{21})\)

(i) \(\sqrt{8\over 18} =\sqrt{4\over 9} = 2/3\) ஒரு விகிதமுறு எண் 
(2) 7/3 ஒரு விகிதமுறு எண் 
(3) \( \sqrt{0.01} = \sqrt{1\over 100} = 1/10\) ஒரு விகிதமுறு எண் 
(4) \( \sqrt{13}\) ஒரு விகிதமுறா எண் 
(∵ முடியுறா சூழல் தன்மைக்கொண்டது)