\(\sqrt{\left(x-\sqrt{5}\right)^2}+\sqrt{\left(y+\sqrt{3}\right)}+\left|x-y-z\right|=0\)

\(\Leftrightarrow\left|x-\sqrt{5}\right|+\left|y+\sqrt{3}\right|+\left|x-y-z\right|=0\)

Ta có \(\hept{\begin{cases}\left|x-\sqrt{5}\right|\ge0\\\left|y+\sqrt{3}\right|\ge0\\\left|x-y-z\right|\ge0\end{cases}}\)

=>  \(VT\ge0\)

Dấu = xảy ra khi

\(\hept{\begin{cases}x-\sqrt{5}=0\\y+\sqrt{3}=0\\x-y-z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\sqrt{5}\\y=-\sqrt{3}\\z=\sqrt{5}+\sqrt{3}\end{cases}}\)

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)

<=>\(\left|x-\sqrt{2}\right|+\left|y+\sqrt{2}\right|+\left|x+y+z\right|=0\)

Vì \(\left|x-\sqrt{2}\right|\ge0;\left|y+\sqrt{2}\right|\ge0;\left|x+y+z\right|\ge0\)

=>\(\left|x-\sqrt{2}\right|+\left|y+\sqrt{2}\right|+\left|x+y+z\right|\ge0\)

Dấu "=" xảy ra khi \(\left|x-\sqrt{2}\right|=\left|y+\sqrt{2}\right|=\left|x+y+z\right|=0\)

\(\left|x-\sqrt{2}\right|=0\Leftrightarrow x-\sqrt{2}=0\Leftrightarrow x=\sqrt{2};\left|y+\sqrt{2}\right|=0\Leftrightarrow y+\sqrt{2}=0\Leftrightarrow y=-\sqrt{2}\)

\(\left|x+y+z\right|=0\Leftrightarrow x+y+z=0\Leftrightarrow\sqrt{2}+\left(-\sqrt{2}\right)+z=0\Leftrightarrow z=0\)

Vậy .......

do căn >= 0 lx+y+zl >=0 nên vế trái >=0
mà vế trái =0 => từng cái =0

4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)

\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)

Tìm z thì dễ rồi

giusp mình nhé mình đang cần lắm

Ta thấy : VT >= 0

Dấu "=" xảy ra <=> x-\(\sqrt{2}\)= 0 ; y+\(\sqrt{2}\)= 0 ; x+y+z = 0 

<=> x=\(\sqrt{2}\);  y=\(-\sqrt{2}\); z = 0

Vậy ...........

Tk mk nha

x=2024, z= 506, y=0