Question

Ofer 25 de puncte :) $\sqrt{x+ \sqrt{2x-1}} + \sqrt{x- \sqrt{2x-1} } = \sqrt{2}$

Answer 1

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

$2x+2\sqrt{(x+ \sqrt{2x-1} )(x- \sqrt{2x}-1 )} =2$
$2x+2 \sqrt{ x^{2} -x \sqrt{2x-1} + x\sqrt{2x-1}-(2x-1)} =2$
$2x+2 \sqrt{ x^{2} +1-2x}=2<=>2x+2 \sqrt{ (x-1)^{2}}=2<=>2x+2(x-1)=2<=>2x-2+2x$=2
<=>4x=2=>x=2

Answer 2

  se ridica la patrat toata relatia si se obtine:
x+$\sqrt{2x-1}$+2$\sqrt{x+ \sqrt{2x-1} } \sqrt{x- \sqrt{2x-1} }$+x-$\sqrt{2x-1}$=2
2x+2$\sqrt{(x+ \sqrt{2x-1)}(x- \sqrt{2x-1}) } }$=2
2x+2$\sqrt{ x^{2} -2x+1}$=2
2x+2$\sqrt{(x-1)^2}$=2
2x+2(x-1)=2
2x+2x-2=2
4x=4
x=1