Question

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Answer 1

Question:-(1)

$\frac { 6 + \sqrt { 2 } } { 2 + \sqrt { 2 } } = x - y$

Answer:-

$\frac { 6 + \sqrt { 2 } } { 2 + \sqrt { 2 } } = x - y \\ \\ = > \frac{\left(6+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=x-y \\ \\ = > \frac{\left(6+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2^{2}-\left(\sqrt{2}\right)^{2}}=x-y \\ \\ = > \frac{\left(6+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=x-y \\ \\ = > \frac{\left(6+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{4-2}=x-y \\ \\ = > \frac{\left(6+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=x-y \\ \\ = > \frac{12-4\sqrt{2}-\left(\sqrt{2}\right)^{2}}{2}=x-y \\ \\ = > \frac{12-4\sqrt{2}-2}{2}=x-y \\ \\ = > \frac{10-4\sqrt{2}}{2}=x-y \\ \\ = > 5-2^{\frac{3}{2}}=x-y \\ \\ = > x-y=5-2^{\frac{3}{2}} \\ \\ = > -y=5-2^{\frac{3}{2}}-x \\ \\ = > \frac{-y}{-1}=\frac{-x+5-2\sqrt{2}}{-1} \\ \\ = > y=\frac{-x+5-2\sqrt{2}}{-1} \\ \\ = > y=x+2\sqrt{2}-5$

Question:-(2)

$\frac { \sqrt { 11 } - \sqrt { 7 } } { \sqrt { 11 } + \sqrt { 7 } } = x - y \sqrt { 77 }$

Answer:-

$\frac { \sqrt { 11 } - \sqrt { 7 } } { \sqrt { 11 } + \sqrt { 7 } } = x - y \sqrt { 77 } \\ \\ = > \frac{\left(\sqrt{11}-\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}+\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)}=x-y\sqrt{77} \\ \\ = > \frac{\left(\sqrt{11}-\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}\right)^{2}-\left(\sqrt{7}\right)^{2}}=x-y\sqrt{77} \\ \\ = > \frac{\left(\sqrt{11}-\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)}{11-7}=x-y\sqrt{77} \\ \\ = > \frac{\left(\sqrt{11}-\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)}{4}=x-y\sqrt{77} \\ \\ = > \frac{\left(\sqrt{11}-\sqrt{7}\right)^{2}}{4}=x-y\sqrt{77} \\ \\ = > \frac{\left(\sqrt{11}\right)^{2}-2\sqrt{11}\sqrt{7}+\left(\sqrt{7}\right)^{2}}{4}=x-y\sqrt{77} \\ \\ = > \frac{11-2\sqrt{11}\sqrt{7}+\left(\sqrt{7}\right)^{2}}{4}=x-y\sqrt{77} \\ \\ = > \frac{11-2\sqrt{77}+\left(\sqrt{7}\right)^{2}}{4}=x-y\sqrt{77} \\ \\ = > \frac{11-2\sqrt{77}+7}{4}=x-y\sqrt{77} \\ \\ = > \frac{18-2\sqrt{77}}{4}=x-y\sqrt{77} \\ \\ = > \frac{9}{2}-\frac{1}{2}\sqrt{77}=x-y\sqrt{77} \\ \\ = > x-y\sqrt{77}=\frac{9}{2}-\frac{1}{2}\sqrt{77} \\ \\ = > -y\sqrt{77}=\frac{9}{2}-\frac{1}{2}\sqrt{77}-x \\ \\ = > \left(-\sqrt{77}\right)y=-x-\frac{\sqrt{77}}{2}+\frac{9}{2} \\ \\ = > \frac{\left(-\sqrt{77}\right)y}{-\sqrt{77}}=\frac{-x-\frac{\sqrt{77}}{2}+\frac{9}{2}}{-\sqrt{77}} \\ \\ = > y=\frac{-x-\frac{\sqrt{77}}{2}+\frac{9}{2}}{-\sqrt{77}} \\ \\ = > y=\frac{\sqrt{77}x}{77}-\frac{9\sqrt{77}}{154}+\frac{1}{2}$

Question:-(3)

$\frac { \sqrt { 7 - 1 } } { \sqrt { 7 + 1 } } - \frac { \sqrt { 7 + 1 } } { \sqrt { 7 - 1 } } = x + y \sqrt { 7 }$

Answer:-

$\frac { \sqrt { 7 - 1 } } { \sqrt { 7 + 1 } } - \frac { \sqrt { 7 + 1 } } { \sqrt { 7 - 1 } } = x + y \sqrt { 7 } \\ \\ = > \frac{\sqrt{6}}{\sqrt{7+1}}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{\sqrt{6}}{\sqrt{8}}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{\sqrt{6}}{2\sqrt{2}}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{\sqrt{6}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{\sqrt{6}\sqrt{2}}{2\times 2}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{\sqrt{2}\sqrt{3}\sqrt{2}}{2\times 2}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{2\sqrt{3}}{2\times 2}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{2\sqrt{3}}{4}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{\sqrt{7+1}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{\sqrt{8}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2\sqrt{2}}{\sqrt{7-1}}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2\sqrt{2}}{\sqrt{6}}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2\sqrt{2}\sqrt{6}}{\left(\sqrt{6}\right)^{2}}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2\sqrt{2}\sqrt{6}}{6}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2\sqrt{2}\sqrt{2}\sqrt{3}}{6}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2\times 2\sqrt{3}}{6}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2\times 2\sqrt{3}}{6}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{4\sqrt{3}}{6}=x+y\sqrt{7} \\ \\ = > \frac{1}{2}\sqrt{3}-\frac{2}{3}\sqrt{3}=x+y\sqrt{7} \\ \\ = > x+y\sqrt{7}=\frac{1}{2}\sqrt{3}-\frac{2}{3}\sqrt{3} \\ \\ = > x=\frac{1}{2}\sqrt{3}-\frac{2}{3}\sqrt{3}-y\sqrt{7} \\ \\ = > x=-\frac{1}{6}\sqrt{3}-y\sqrt{7}$

Answer 2

thank you so much........ dear.... ❤