Question

maths
solve and answer me ​

Answer 1

Answer:

$1) let \: x = 2 + \sqrt{3} \\ value \: of \: {x}^{2} + \frac{1}{ {x}^{2} } \\ substituting \: value \: we \: get \\ {(2 + \sqrt{3} ) }^{2} + \frac{1}{(2 + { \sqrt{3)} }^{2} } \\ 4 + 3 \: + \frac{1}{4 + 3} \\ 7 + \frac{1}{7} \\ \frac{49 + 1}{7} = \frac{50}{7}$

2) the value of √20 ×√5 = √100 = 10

Answer 2

Remember, it is your punishment for spamming in my question.

Given:

$\bf\displaystyle \int^4_2\frac{\sqrt{x} }{\sqrt{6 - x + x} } dx$

To find : Integral

Explanation:

$\bf\displaystyle \int^4_2\frac{\sqrt{x} }{\sqrt{6 } } dx$

$\bf\displaystyle \frac{1}{\sqrt{6 } }\int^4_2\sqrt{x} dx$

Do integration

$\int {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} + c$

$=\bf\displaystyle\frac{2}{\sqrt{6} } [\frac{x^{\frac{3}{2}}}{3}]$

put lower and upper limits

$= \frac{2}{ \sqrt{6} } ( \frac{4 \times 2}{3} - \frac{2 \sqrt{2} }{3} ) \\ \\ = \frac{2}{3 \sqrt{6} } (8 - 2\sqrt{2} )$

or

$= \frac{2}{ \sqrt{6} } ( \frac{4 \times 2}{3} - \frac{2 \sqrt{2} }{3} ) \\ \\ = \frac{4}{3 \sqrt{6} } (4 - \sqrt{2} )$

Thus,

$\bf\displaystyle \int^4_2\frac{\sqrt{x} }{\sqrt{6 - x + x} } dx=\frac{4}{3 \sqrt{6} } (4 - \sqrt{2} )$

Hope it helps you.