Câu hỏi của Nguyễn Tuấn Kiệt

Question

Giải pt sau:

1/ $\sqrt{4x^2-12x+9}=3-2x$

2/ $\sqrt{x^2-2\sqrt{2}x+2}=\sqrt{9-4\sqrt{2}}-\sqrt{3+2\sqrt{2}}$

Nguyễn Lê Phước Thịnh · Accepted Answer

1: Ta có: $\sqrt{4x^2-12x+9}=3-2x$

$\Leftrightarrow\left(2x-3\right)^2=\left(3-2x\right)^2$

$\Leftrightarrow\left(2x-3\right)^2-\left(3-2x\right)^2=0$

$\Leftrightarrow\left[\left(2x-3\right)-\left(3-2x\right)\right]\left[\left(2x-3\right)+\left(3-2x\right)\right]=0$

$\Leftrightarrow\left(2x-3-3+2x\right)\left(2x-3+3-2x\right)=0$

$\Leftrightarrow\left(4x-6\right)\cdot0=0$(luôn đúng)

Vậy: S={x|$x\in R$}

2) Ta có: $\sqrt{x^2-2\cdot\sqrt{2}\cdot x+2}=\sqrt{9-4\sqrt{2}}-\sqrt{3+2\sqrt{2}}$

$\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8-2\cdot2\sqrt{2}\cdot1+1}-\sqrt{1+2\cdot1\cdot\sqrt{2}+2}$

$\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\left|\sqrt{8}-1\right|-\left|1+\sqrt{2}\right|$

$\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8}-1-1-\sqrt{2}$

$\Leftrightarrow\left|x-\sqrt{2}\right|=\sqrt{2}-2$(*)

Trường hợp 1: $x\ge\sqrt{2}$

(*)$\Leftrightarrow x-\sqrt{2}=\sqrt{2}-2$

$\Leftrightarrow x-\sqrt{2}-\sqrt{2}+2=0$

$\Leftrightarrow x-2\sqrt{2}+2=0$

$\Leftrightarrow x=2\sqrt{2}-2$(loại)

Trường hợp 2: $x< \sqrt{2}$

(*)$\Leftrightarrow\sqrt{2}-x=\sqrt{2}-2$

$\Leftrightarrow\sqrt{2}-x-\sqrt{2}+2=0$

$\Leftrightarrow2-x=0$

hay x=2(loại)

Vậy: S=∅Câu trả lời: 1: Ta có: $\sqrt{4x^2-12x+9}=3-2x$