Question

Answer 1

a.

ĐKXĐ: \(\left\{\begin{matrix} x-5\geq 0\\ 5-x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ x\leq 5\end{matrix}\right.\Rightarrow x=5\)

Thay $x=5$ vô pt ban đầu:

$0+0=1$ (vô lý)

Vậy pt vô nghiệm

b. Tương tự câu a, nhưng thay $x=\frac{3}{2}$ vào pt thấy thỏa mãn

Do đó pt có nghiệm $x=\frac{3}{2}$

c. ĐKXĐ: $x=-1$ hoặc $x\geq 1$

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

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

Vì $3\sqrt{x-1}+2\geq 2>0$ nên $\sqrt{x+1}=0$

$\Leftrightarrow x=-1$

 

 

Answer 2

d. ĐKXĐ: $x\geq -3; y\geq 2; z\geq 4$
PT $\Leftrightarrow 2\sqrt{x+3}+2\sqrt{y-3}+2\sqrt{z-4}=x+y+z$

$\Leftrightarrow [(x+3)-2\sqrt{x+3}+1]+[(y-3)-2\sqrt{y-3}+1]+[(z-4)-2\sqrt{z-4}+1]=0$

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

$\Rightarrow \sqrt{x+3}-1=\sqrt{y-3}-1=\sqrt{z-4}-1=0$

$\Leftrightarrow x=-2; y=4; z=5$

Answer 3

e. ĐKXĐ: $x\geq 0; y\geq 1$

PT $\Leftrightarrow (x-2\sqrt{x}+1)+(y-1-4\sqrt{y-1}+4)=0$

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

$\Rightarrow \sqrt{x}-1=\sqrt{y-1}-2=0$

$\Leftrightarrow x=1; y=5$

f.

PT$\Leftrightarrow \sqrt{x}+2\sqrt{y+1}+\sqrt{y}=2\sqrt{y+1}$

$\Leftrightarrow \sqrt{x}+\sqrt{y}=0$

$\Rightarrow x=y=0$

g. ĐKXĐ: $x\geq -1; y\geq 3; z\geq 1$

PT $\Leftrightarrow 2\sqrt{x+1}+2\sqrt{y-3}+2\sqrt{z-1}=x+y+z$

$\Leftrightarrow [(x+1)-2\sqrt{x+1}+1]+[(y-3)-2\sqrt{y-3}+1]+[(z-1)-2\sqrt{z-1}+1]=0$

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

$\Rightarrow \sqrt{x+1}-1=\sqrt{y-3}-1=\sqrt{z-1}-1=0$

$\Leftrightarrow x=0; y=4; z=2$