a) \(\left\{{}\begin{matrix}x+3y+2z=8\\2x+2y+z=6\\3x+y+z=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=2\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}x-3y+2z=-7\\-2x+4y+3z=8\\3x+y-z=5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{11}{14}\\y=\dfrac{5}{2}\\z=-\dfrac{1}{7}\end{matrix}\right.\)

a) Đặt \(\left\{{}\begin{matrix}x+3y+2z=8\left(1\right)\\2x+2y+z=6\left(2\right)\\3x+y+z=6\left(3\right)\end{matrix}\right.\)
Cộng \(\left(2\right)+\left(3\right)\) ta có:\(\left\{{}\begin{matrix}x+3y+2z=8\left(1\right)\\2x+2y+z=6\left(2\right)\\5x+3y+2z=12\left(4\right)\end{matrix}\right.\)
Trừ \(\left(4\right)-\left(1\right)\) ta được: \(4x=4\Leftrightarrow x=1\).
Thay vào hệ phương trình ta được:
\(\left\{{}\begin{matrix}1+3y+2z=8\\2.1+2y+z=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\z=2\end{matrix}\right.\).
Vậy hệ phương trình có nghiệm: \(\left\{{}\begin{matrix}x=1\\y=1\\z=2\end{matrix}\right.\).

b) Đặt \(\left\{{}\begin{matrix}x-3y+2z=-7\left(1\right)\\-2x+4y+3z=8\left(2\right)\\3x+y-z=5\left(3\right)\end{matrix}\right.\)
Cộng \(\left(1\right)-\left(2\right)\) ta được: \(3x-7y-z=-15\left(4\right)\)
Lấy \(\left(3\right)-\left(4\right)\) ta được: \(8y=20\Leftrightarrow y=\dfrac{5}{2}\).
Thay \(y=\dfrac{5}{2}\) vào hệ phương trình ta có:
\(\left\{{}\begin{matrix}x-3.\dfrac{5}{2}+2z=-7\\-2x+4.\dfrac{5}{2}+3z=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{14}\\z=-\dfrac{1}{7}\end{matrix}\right.\).
Vậy hệ có nghiệm là: \(\left\{{}\begin{matrix}x=\dfrac{11}{14}\\y=\dfrac{5}{2}\\z=\dfrac{-1}{7}\end{matrix}\right.\)

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

\(x^3=3y^2-3y+1=3\left(y-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(\Rightarrow x\ge\dfrac{1}{\sqrt[3]{4}}>\dfrac{1}{2}\)

Tương tự ta có \(y;z>\dfrac{1}{2}\)

\(\Rightarrow x+y-1>0;y+z-1>0;z+x-1>0\)

TH1: \(x\ge y\Rightarrow x^3\ge y^3\Rightarrow3y^2-3y+1\ge3z^2-3z+1\)

\(\Rightarrow y^2-z^2-y+z\ge0\Rightarrow\left(y-z\right)\left(y+z+1\right)\ge0\)

\(\Rightarrow y-z\ge0\Rightarrow y\ge z\Rightarrow x\ge z\) (1)

Cũng do \(y\ge z\Rightarrow y^3\ge z^3\)

\(\Rightarrow3z^2-3z+1\ge3x^2-3x+1\Rightarrow z^2-x^2-z+x\ge0\)

\(\Rightarrow\left(z-x\right)\left(z+x+1\right)\ge0\Rightarrow z\ge x\) (2)

Từ (1);(2) \(\Rightarrow x=y=z\)

TH2: \(x\le y\), hoàn toàn tương tự ta cũng chứng minh được \(x=y=z\)

Thay vào hệ ban đầu:

\(\left\{{}\begin{matrix}x^3-3x^2+3x=1\\y^3-3y^2+3y=1\\z^3-3z^2+3z=1\end{matrix}\right.\) \(\Rightarrow x=y=z=1\)

x=y=z=2

b) Đặt \(\left\{{}\begin{matrix}x+y+z=7\left(1\right)\\3x-2y+2z=5\left(2\right)\\4x-y+3z=10\left(3\right)\end{matrix}\right.\)
Cộng \(\left(1\right)+\left(2\right)\) ta có: \(4x-y+3z=12\). (4)
Từ (3) và (4): \(\left\{{}\begin{matrix}4x-y+3z=12\\4x-y+3z=10\end{matrix}\right.\) (vô nghiệm).
Vậy hệ phương trình vô nghiệm.

a. \(\left\{{}\begin{matrix}3x-5y=6\\4x+7y=-8\end{matrix}\right.\)

\(x=\dfrac{2}{41}\) ; \(y=\dfrac{-48}{41}\)

b.

\(x=\dfrac{2}{19};y=\dfrac{33}{19}\)

c.

\(x=\dfrac{22}{101};y=\dfrac{131}{101};z=\dfrac{-39}{101}\)

d.

\(x=-4;y=\dfrac{11}{7};z=\dfrac{12}{7}\)

a)x=0,05 ; y=-1,17

b.x=0,11 ; y=1,74

c.x=0,22 ;y=1,29 z=-0.39

d.x=-4 y=1,57 z=1,71

a,\(\left\{{}\begin{matrix}3x-5y=6\\4x+7y=-8\end{matrix}\right.\)

x=\(\dfrac{2}{41}=0,05\) ; y=\(\dfrac{-48}{41}=-1,17\)

b,\(\left\{{}\begin{matrix}-2x+3y=5\\5x+2y=4\end{matrix}\right.\)

x=\(\dfrac{2}{19}=0,11\) ; y=\(\dfrac{33}{19}=1,74\)

c,\(\left\{{}\begin{matrix}2x-3y+4z=-5\\-4x+5y-z=6\\3x+4y-3z=2\end{matrix}\right.\)

x=\(\dfrac{22}{101}=0,22\) ;y=\(\dfrac{131}{101}=1,29\) ; z=\(\dfrac{-39}{101}=-0,39\)

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

x=\(-4\) ; y=\(\dfrac{11}{7}=1,57\) ; z=\(\dfrac{12}{7}=1,71\)

Không mất tính tổng quát, giả sử \(x=min\left\{x;y;z\right\}\)

\(\Rightarrow z=3x^3+2x^2+x\le3y^3+2y^2+y\)

\(\Rightarrow z\le x\)

\(\Rightarrow z=x\)

\(\Rightarrow x=y=z\)

\(\Rightarrow x=3y^3+2x^2+x\Rightarrow x^2\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\end{matrix}\right.\)

@Nguyễn Việt Lâm