Question

$\sqrt{7-x}=1$
$\sqrt{2+x}=x$$\sqrt{2x+x}=x$

Answer 1

1.($(\sqrt{7-x} )^{2} = 1^{2} => -x=1-7 => x=6$
2. a) $( \sqrt{2+x})^{2} = x^{2} => 2+x- x^{2} =0 /*(-1) => x^{2} -x-2=0$
Δ$= b^{2} -4ac=> 1+8=> 9$
[tex] x_{1,2} = \frac{-b+- \sqrt{delta} }{2a} => x_{1,2} = \frac{1+-3}{2} => x_{1} =2 ; x_{2} = -1[/tex]
b) [tex] \sqrt{3x ^{3} } =x => {3x^{3} } = x^{2} => 3x^{3} - x^{2} =0; x^{2} (3x-1)=0 => \left \{ {{ x_{1} =o} \atop { x_{2}= \frac{1}{3} }} \right. [/tex]