Question

2 by 3 root 7 minus 1 by 2 root 2 + 6 root 11 and 1 by 3 root 7 + 312 root 2 minus root 11

Answer 1

Hope it will help you plzz mark it

Answer 2

Answer:

The correct answer is $&\sqrt{7}+\sqrt{2}+5 \sqrt{11}$.

Step-by-step explanation:

Given:

$\left(\frac{2}{3} \sqrt{7}-\frac{1}{2} \sqrt{2}+6 \sqrt{11}\right)$ ...............(1)

and $\left(\frac{1}{3} \sqrt{7}+\frac{3}{2} \sqrt{2}-\sqrt{11}\right)$..........(2)

Step 1

Given:

$\left(\frac{2}{3} \sqrt{7}-\frac{1}{2} \sqrt{2}+6 \sqrt{11}\right)$ ...............(1)

and $\left(\frac{1}{3} \sqrt{7}+\frac{3}{2} \sqrt{2}-\sqrt{11}\right)$..........(2)

Adding (1) and (2), then we get

$&\left(\frac{2}{2} \sqrt{7}-\frac{1}{2} \sqrt{2}+6 \sqrt{11}\right)+\left(\frac{1}{3} \sqrt{7}+\frac{3}{2} \sqrt{2}-\sqrt{11}\right)$

Step 2

Simplifying the above equation, we get

$&=\left(\frac{2}{3} \sqrt{7}+\frac{1}{3} \sqrt{7}\right)+\left(\frac{-1}{2} \sqrt{2}+\frac{3}{2} \sqrt{2}\right)+(6 \sqrt{11}-\sqrt{11})$

equating

$&=\left(\frac{2}{3}+\frac{1}{3}\right) \sqrt{7}+\left(\frac{3}{2}-\frac{1}{2}\right) \sqrt{2}+(6-1) \sqrt{11}$

$&=\frac{3}{3} \sqrt{7}+\frac{2}{2} \sqrt{2}+5(\sqrt{11})$

$&=\sqrt{7}+\sqrt{2}+5 \sqrt{11}$

we get $&\sqrt{7}+\sqrt{2}+5 \sqrt{11}$.

Therefore, the correct answer is $&\sqrt{7}+\sqrt{2}+5 \sqrt{11}$.

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