Question

[Nhân Nguyễn]

\(\sqrt{x}+\sqrt{2-x}+\sqrt{x\left(2-x\right)}=\sqrt{2}\)

Answer 1

[Akai Haruma]

Lời giải:
ĐKXĐ: $0\leq x\leq 2$

Đặt $\sqrt{x}=a; \sqrt{2-x}=b(a,b\geq 0)$

$\Rightarrow a^2+b^2=2$

$\Leftrightarrow (a+b)^2-2ab=2(1)$
PT đã cho trở thành:

$a+b+ab=\sqrt{2}$

$\Leftrightarrow ab=\sqrt{2}-(a+b)$. Thay vào $(1)$:

$(a+b)^2-2[\sqrt{2}-(a+b)]=2$

$\Leftrightarrow (a+b)^2+2(a+b)=2+2\sqrt{2}$

$\Leftrightarrow (a+b+1)^2=3+2\sqrt{2}$

$\Rightarrow a+b+1=\sqrt{2}+1$ hoặc $a+b+1=-(\sqrt{2}+1)$

Vì $a,b\geq 0$ nên $a+b+1=\sqrt{2}+1$
$\Leftrightarrow a+b=\sqrt{2}$

$ab=\sqrt{2}-(a+b)=\sqrt{2}-\sqrt{2}=0$

Vậy $(a+b, ab) = (\sqrt{2},0)$

$\Rightarrow (a,b)=(0,\sqrt{2}), (\sqrt{2},0)$

$\Rightarrow (\sqrt{x},\sqrt{2-x})=(0,\sqrt{2}), (\sqrt{2},0)$

$\Rightarrow x=0$ hoặc $x=2$