Giúp em với ạ! Em cảm ơn nhiềuuu

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)

chả cần HĐT dùng Cosi cx đc

\(\left(x+1\right)+4\ge4\sqrt{x+1}\)

\(\left(y+2\right)+9\ge6\sqrt{y+2}\)

\(\left(z+3\right)+16\ge8\sqrt{z+3}\)

\(\Rightarrow VT\ge VP\).Dấu = khi x=3;y=7;z=13

\(ĐKXĐ:\left\{{}\begin{matrix}x\ge-1\\y\ge-2\\z\ge-3\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ge0\\y+2\ge0\\z+3\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-1\\y\ge-2\\z\ge-3\end{matrix}\right.\)

Ta có : \(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)

=> \(x+y+z+35=4\sqrt{x+1}+6\sqrt{y+2}+8\sqrt{z+3}\)

=> \(x-4\sqrt{x+1}+y-6\sqrt{y+2}+z-8\sqrt{z+3}+35=0\)

=> \(x+1-2.2\sqrt{x+1}+4+y+2-2.3\sqrt{y+2}+9+z+3-4.2\sqrt{z+3}+16=0\)

=> \(\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)

Ta thấy : \(\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2\ge0\\\left(\sqrt{y+2}-3\right)^2\ge0\\\left(\sqrt{z+3}-4\right)^2\ge0\end{matrix}\right.\)

=> \(\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2\ge0\)

- Dấu "=" xảy ra

<=> \(\left\{{}\begin{matrix}\sqrt{x+1}-2=0\\\sqrt{y+2}-3=0\\\sqrt{z+3}-4=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x+1=4\\y+2=9\\z+3=16\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\) ( TM )

Vậy ...

\(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)

\(\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0\)

\(\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(y+2-6\sqrt{y+2}+9\right)+\left(z+3-8\sqrt{z+3}+16\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(\sqrt{y+2}-3\right)^2=0\\\left(\sqrt{z+3}-4\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=7\\z=13\end{cases}}\)

ĐKXĐ: ....

\(x+1-4\sqrt{x+1}+4+y+2-6\sqrt{y+2}+9+z+3-8\sqrt{z+3}+16=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\)