Question

pls help me please pls​

Answer 1

Answer:

Therefore, x2+ 1x2 1 x 2 = 47

Answer 2

Given :-

$\bigg\lgroup \sf x+\dfrac{1}{x}\bigg\rgroup=7$

To Find :-

$\bigg\lgroup \sf x^2+\dfrac{1}{x^2}\bigg\rgroup$

Solution :-

$\bigg\lgroup \sf x+\dfrac{1}{x}\bigg\rgroup=7$

On squaring both sides

$\bigg\lgroup \sf x+\dfrac{1}{x}\bigg\rgroup^2=(7)^2$

We know that

$\sf (x+y)^2=x^2+2xy+y^2$

$\sf x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}=49$

$\sf x^2+\dfrac{1}{x^2}+2=49$

$\sf x^2+\dfrac{1}{x^2}=49-2$

${\pmb{\bf x^2+\dfrac{1}{x^2}=47}}$