Câu hỏi của Trần Hoàng Đạt

Question

Giải phương trình vô tỉ:

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

b) $\sqrt{3x+8+6\sqrt{3x-1}}+\sqrt{3x+8-6\sqrt{3x-1}}=3x+4$

c) $2x\sqrt{x^2-x-1}+4x\sqrt{3x+1}=2x^2+2x+6$

Akai Haruma · Accepted Answer

a) ĐKXĐ: $-1\leq x\leq 2$

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

$\Leftrightarrow \sqrt{2+x-x^2}=1+2x-2x^2=-3+2(2+x-x^2)$

Đặt $\sqrt{2+x-x^2}=t(t\geq 0)$. PT trở thành:

$t=-3+2t^2$

$\Leftrightarrow 2t^2-t-3=0\Leftrightarrow (2t-3)(t+1)=0$

$\Rightarrow t=\frac{3}{2}$ (do $t\geq 0)$

$\Rightarrow 2+x-x^2=\frac{9}{4}\Rightarrow x^2-x+\frac{1}{4}=0$

$\Leftrightarrow (x-\frac{1}{2})^2=0\Rightarrow x=\frac{1}{2}$ (thỏa mãn)Câu trả lời: a) ĐKXĐ: $-1\leq x\leq 2$

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

$\Leftrightarrow \sqrt{2+x-x^2}=1+2x-2x^2=-3+2(2+x-x^2)$

Đặt $\sqrt{2+x-x^2}=t(t\geq 0)$. PT trở thành:

$t=-3+2t^2$

$\Leftrightarrow 2t^2-t-3=0\Leftrightarrow (2t-3)(t+1)=0$

$\Rightarrow t=\frac{3}{2}$ (do $t\geq 0)$

$\Rightarrow 2+x-x^2=\frac{9}{4}\Rightarrow x^2-x+\frac{1}{4}=0$

$\Leftrightarrow (x-\frac{1}{2})^2=0\Rightarrow x=\frac{1}{2}$ (thỏa mãn)

Akai Haruma · Answer

b) ĐK: $x\geq \frac{1}{3}$

PT $\Leftrightarrow \sqrt{(3x-1)+6\sqrt{3x-1}+9}+\sqrt{(3x-1)-6\sqrt{3x-1}+9}=3x+4$

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

$\Leftrightarrow \sqrt{3x-1}+3+|\sqrt{3x-1}-3|=3x+4$

$\Leftrightarrow |\sqrt{3x-1}-3|=3x-\sqrt{3x-1}+1$

Nếu $\sqrt{3x-1}\geq 3$:

$\Rightarrow \sqrt{3x-1}-3=3x-\sqrt{3x-1}+1$

$\Leftrightarrow 3x+4-2\sqrt{3x-1}=0$

$\Leftrightarrow (3x-1)-2\sqrt{3x-1}+5=0$

$\Leftrightarrow (\sqrt{3x-1}-1)^2+4=0$ (vô lý)

Nếu $\sqrt{3x-1}< 3$:

$\Rightarrow 3-\sqrt{3x-1}=3x-\sqrt{3x-1}+1$

$\Leftrightarrow 3x=2\Rightarrow x=\frac{2}{3}$ (thỏa mãn)

Vậy...........Câu trả lời: b) ĐK: $x\geq \frac{1}{3}$