5−75+7−5+75−7\dfrac{5 - \sqrt{7}}{5 + \sqrt{7}} - \dfrac{5 + \sqrt{7}}{5 - \sqrt{7}}5+75−7−5−75+7 is equal to :
10710\sqrt{7}107
197\dfrac{1}{9}\sqrt{7}917
1079\dfrac{10\sqrt{7}}{9}9107
−1079-\dfrac{10\sqrt{7}}{9}−9107
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Given,
⇒5−75+7−5+75−7⇒(5−7)2−(5+7)2(5+7)(5−7)⇒(5)2+(7)2−2×5×7−[(5)2+(7)2+2×5×7]52−(7)2⇒25+7−107−[25+7+107]25−7⇒25−25+7−7−107−10718⇒−20718⇒−1079.\Rightarrow \dfrac{5 - \sqrt{7}}{5 + \sqrt{7}} - \dfrac{5 + \sqrt{7}}{5 - \sqrt{7}} \\[1em] \Rightarrow \dfrac{(5 - \sqrt{7})^2 - (5 + \sqrt{7})^2}{(5 + \sqrt{7})(5 - \sqrt{7})} \\[1em] \Rightarrow \dfrac{(5)^2 + (\sqrt{7})^2 - 2 \times 5 \times \sqrt{7} - [(5)^2 + (\sqrt{7})^2 + 2 \times 5 \times \sqrt{7}]}{5^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \dfrac{25 + 7 - 10\sqrt{7} - [25 + 7 + 10\sqrt{7}]}{25 - 7} \\[1em] \Rightarrow \dfrac{25 - 25 + 7 - 7 - 10\sqrt{7} - 10\sqrt{7}}{18} \\[1em] \Rightarrow \dfrac{-20\sqrt{7}}{18} \\[1em] \Rightarrow \dfrac{-10\sqrt{7}}{9}.⇒5+75−7−5−75+7⇒(5+7)(5−7)(5−7)2−(5+7)2⇒52−(7)2(5)2+(7)2−2×5×7−[(5)2+(7)2+2×5×7]⇒25−725+7−107−[25+7+107]⇒1825−25+7−7−107−107⇒18−207⇒9−107.
Hence, Option 4 is the correct option.
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17−5\dfrac{1}{7 - \sqrt{5}}7−51 is equal to :
4(7+5)4(7 + \sqrt{5})4(7+5)
144(7+5)\dfrac{1}{44}(7 + \sqrt{5})441(7+5)
144(7−5)\dfrac{1}{44}(7 - \sqrt{5})441(7−5)
4(7−5)4(7 - \sqrt{5})4(7−5)
If x = 2−1, then (x−1x)2\sqrt{2} - 1, \text{ then } \Big(x - \dfrac{1}{x}\Big)^22−1, then (x−x1)2 is :
222\sqrt{2}22
2
4
2−22 - \sqrt{2}2−2
State, with reason, which of the following are surds and which are not :
(i) 180\sqrt{180}180
(ii) 274\sqrt[4]{27}427
(iii) 1285\sqrt[5]{128}5128
(iv) 643\sqrt[3]{64}364
(v) 253.403\sqrt[3]{25}.\sqrt[3]{40}325.340
(vi) −1253\sqrt[3]{-125}3−125
(vii) π\sqrt{π}π
(viii) 3+2\sqrt{3 + \sqrt{2}}3+2
Write the lowest rationalizing factor of :
(i) 525\sqrt{2}52
(ii) 24\sqrt{24}24
(iii) 5−3\sqrt{5} - 35−3
(iv) 7−77 - \sqrt{7}7−7
(v) 18−50\sqrt{18} - \sqrt{50}18−50
(vi) 5−2\sqrt{5} - \sqrt{2}5−2
(vii) 13+3\sqrt{13} + 313+3
