Question

[Thu Maii Nguyễn]

Giúp mình bài này với ạ, ghi rõ cả dkxd ra nha

Answer 1

[Akai Haruma]

a. ĐKXĐ: $x\in\mathbb{R}$

PT \(\Rightarrow \left\{\begin{matrix} 2-x\geq 0\\ x^2+x+2=(3-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ x^2+x+2=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ 7x=7\end{matrix}\right.\Leftrightarrow x=1\)

b. ĐKXĐ: $x\geq -1$

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

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

$\Leftrightarrow \sqrt{x+1}[(x-1)\sqrt{x+1}+1]=0$

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

Nếu $\sqrt{x+1}=0$

$\Leftrightarrow x=-1$ (tm)

Nếu $(x-1)\sqrt{x+1}+1=0$

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

$\Rightarrow (x-1)^2(x+1)=1$

$\Leftrightarrow x^3-x^2-x=0$

$\Leftrightarrow x(x^2-x-1)=0$

$\Leftrightarrow x=0$ hoặc $x^2-x-1=0$

$\Leftrightarrow x=0$ hoặc $x=\frac{1\pm \sqrt{5}}{2}$

Kết hợp đkxđ suy ra $x=0; -1; \frac{1\pm \sqrt{5}}{2}$

Answer 2

[Akai Haruma]

c. ĐKXĐ: $x\geq 2$

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

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

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

$\Leftrightarrow x=2$  (thỏa mãn)

d. ĐKXĐ: $x\geq 3$ hoặc $x\leq -4$

PT \(\Rightarrow \left\{\begin{matrix} 8-x\geq 0\\ x^2+x-12=(8-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 8\\ x^2+x-12=x^2-16x+64\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 8\\ 17x=76\end{matrix}\right.\Leftrightarrow x=\frac{76}{17}\) (tm)