a.
ĐKXĐ: $x\geq 0; y\geq 1$
PT $\Leftrightarrow (x-4\sqrt{x}+4)+(y-1-6\sqrt{y-1}+9)=0$
$\Leftrightarrow (\sqrt{x}-2)^2+(\sqrt{y-1}-3)^2=0$
Vì $(\sqrt{x}-2)^2; (\sqrt{y-1}-3)^2\geq 0$ với mọi $x\geq 0; y\geq 1$ nên để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{y-1}-3=0$
$\Leftrightarrow x=4; y=10$
b.
ĐKXĐ: $x\geq -1; y\geq -2; z\geq -3$
PT $\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0$
$\Leftrightarrow [(x+1)-4\sqrt{x+1}+4]+[(y+2)-6\sqrt{y+2}+9]+[(z+3)-8\sqrt{z+3}+16]=0$
$\Leftrightarrow (\sqrt{x+1}-2)^2+(\sqrt{y+2}-3)^2+(\sqrt{z+3}-4)^2=0$
$\Rightarrow \sqrt{x+1}-2=\sqrt{y+2}-3=\sqrt{z+3}-4=0$
$\Rightarrow x=3; y=7; z=13$
c.
ĐKXĐ: $x\geq \frac{-1}{8}$
PT $\Leftrightarrow 9x+17-6\sqrt{8x+1}-4\sqrt{x+3}=0$
$\Leftrightarrow [(8x+1)-6\sqrt{8x+1}+9]+[(x+3)-4\sqrt{x+3}+4]=0$
$\Leftrightarrow (\sqrt{8x+1}-3)^2+(\sqrt{x+3}-2)^2=0$
$\Rightarrow \sqrt{8x+1}-3=\sqrt{x+3}-2=0$
$\Rightarrow x=1$ (thỏa mãn đkxđ)
a) \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\) (ĐK: \(x\ge0;y\ge1\))
<=> \(\left(x-4\sqrt{x}+4\right)+\left(y-1\right)-6\sqrt{y-1}+9=0\)
<=> \(\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)
Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2\ge0\\\left(\sqrt{y-1}-3\right)^2\ge0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{y-1}-3=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=4\left(c\right)\\y=10\left(c\right)\end{matrix}\right.\)
KL: Phương trình có nghiệm (x;y) = (4;10)
d.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 2\sqrt{x}+4\sqrt{x+3}=2x+8$
$\Leftrightarrow [(x+3)-4\sqrt{x+3}+4]+(x-2\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x+3}-2)^2+(\sqrt{x}-1)^2=0$
$\Rightarrow \sqrt{x+3}-2=\sqrt{x}-1=0$
$\Rightarrow x=1$ (thỏa mãn)
e.
ĐKXĐ: $3\geq x\geq \frac{2}{3}$
PT $\Leftrightarrow 2\sqrt{3-x}+4\sqrt{3x-2}-6=2x$
$\Leftrightarrow 2x+6-2\sqrt{3-x}-4\sqrt{3x-2}=0$
$\Leftrightarrow [(3x-2)-4\sqrt{3x-2}+4]+[(3-x)-2\sqrt{3-x}+1]=0$
$\Leftrightarrow (\sqrt{3x-2}-2)^2+(\sqrt{3-x}-1)^2=0$
$\Rightarrow \sqrt{3x-2}-2=\sqrt{3-x}-1=0$
$\Rightarrow x=2$ (thỏa mãn đkxđ)
