Question

giải các pt sau:

1).\(\left(x^2-x+\sqrt{2}-2\right)\sqrt{1-2x}=0\)

2).\(4\sqrt{1+x}-1=3x+2\sqrt{1-x}+\sqrt{1-x^2}\)

Answer 1

1. ĐKXĐ: $x\leq \frac{1}{2}$
PT \(\Leftrightarrow [(x^2-2)-(x-\sqrt{2})]\sqrt{1-2x}=0\)

\(\Leftrightarrow (x-\sqrt{2})(x+\sqrt{2}-1)\sqrt{1-2x}=0\)

\(\Leftrightarrow \left[\begin{matrix} x-\sqrt{2}=0\\ x+\sqrt{2}-1=0\\ \sqrt{1-2x}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\sqrt{2}\\ x=1-\sqrt{2}\\ x=\frac{1}{2}\end{matrix}\right.\)

Kết hợp đkxđ suy ra \(\left[\begin{matrix} x=1-\sqrt{2}\\ x=\frac{1}{2}\end{matrix}\right.\)

Answer 2

2. ĐKXĐ: $-1\leq x\leq 1$

Đặt $\sqrt{1+x}=a; \sqrt{1-x}=b(a,b\geq 0)$. Khi đó ta có:

$4a-\frac{a^2+b^2}{2}=\frac{3(a^2-b^2)}{2}+2b+ab=0$

$\Leftrightarrow 2a^2-b^2+ab-4a+2b=0$

$\Leftrightarrow (a+b-2)(2a-b)=0$

Xét 2 TH:

TH1: $a+b-2=0$

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

$\Leftrightarrow 2+2\sqrt{1-x^2}=4$
$\Leftrightarrow \sqrt{1-x^2}=1$

$\Leftrightarrow x=0$ (tm)

TH2: $2a-b=0$

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

$\Leftrightarrow 4(x+1)=1-x$

$\Leftrightarrow x=\frac{-3}{5}$ (tm)

Vậy.........