X = 96

Y= 80

Z= 48

\(\left\{{}\begin{matrix}x+y+z=224\\-5x+3y+5z=0\\x-2z=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+3y+3z=672\left(1\right)\\-5x+3y+5z=0\left(2\right)\\x-2z=0\left(3\right)\end{matrix}\right.\)

\(\left(1\right)-\left(2\right)\Leftrightarrow8x-2z=672\)

\(\Leftrightarrow4x-z=336\left(4\right)\)

\(\left(3\right);\left(4\right)\Leftrightarrow\left\{{}\begin{matrix}x-2z=0\\4x-z=336\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-8z=0\\4x-z=336\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7z=336\\x-2z=0\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x=96\\z=48\end{matrix}\right.\)

\(\Rightarrow y=224-96-48=80\)

Vậy nghiệm hpt đã cho là \(\left\{{}\begin{matrix}x=96\\y=80\\z=48\end{matrix}\right.\)

pt (2) - pt (3) ta có

\(-3y=2\Leftrightarrow y=\frac{-2}{3}\)

pt (1) - pt (2) - pt (3) ta có

\(2x-14y=-4\Leftrightarrow2x-14\cdot\frac{-2}{3}=-4\Leftrightarrow x=-\frac{20}{3}\)

thay các giá trị x , y vào pt 1 ta có

\(4\cdot\frac{-20}{3}+\frac{2}{3}+4z=0\Leftrightarrow z=\frac{-26}{4}\)

vậy (x,y,z) = \(\left(-\frac{20}{3};-\frac{2}{3};-\frac{26}{4}\right)\)

Có: \(\left\{{}\begin{matrix}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{matrix}\right.\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0^{\left(1\right)}\)

Lại có:

 \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{matrix}\right.\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2\ge0\forall x;y;z^{\left(2\right)}\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x+1=0\\y+1=0\\z+1=0\end{matrix}\right.\Rightarrow x=y=z=-1\)

Thay \(x=y=z=-1\) vào \(A\), ta được:

\(A=\left(-1\right)^{2000}+\left(-1\right)^{2000}+\left(-1\right)^{2000}\)

\(=1+1+1=3\)

\(\text{#}\mathit{Toru}\)

a) \(\left\{{}\begin{matrix}x+2y-3z=2\\2x+7y+z=5\\-3x+3y-2z=-7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2y-3z=2\\3y+7z=1\\-32z=-4\end{matrix}\right.\)

Đáp số : \(\left(x,y,z\right)=\left(\dfrac{55}{24},\dfrac{1}{24},\dfrac{1}{8}\right)\)

b) \(\left\{{}\begin{matrix}-x-3y+4z=3\\3x+4y-2z=5\\2x+y+2z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-3y+4z=3\\-5y+10z=14\\-5y+10z=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-3y+4z=3\\-5y+10z=14\\0y+0z=-4\end{matrix}\right.\)

Phương trình cuối vô nghiệm, suy ra hệ phương trình đã cho vô nghiệm