\(\left\{{}\begin{matrix}x+y+z=11\left(1\right)\\2x-y+z=5\left(2\right)\\3x+2y+z=14\left(3\right)\end{matrix}\right.\)

Từ (1) ta có \(z=11-x-y\)

Thay vào (2) và (3) ta được hệ phương trình:

\(\left\{{}\begin{matrix}2x-y+11-x-y=5\\3x+2y+11-x-y=14\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=-6\\2x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=-6\\4x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=-6\\5x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{x+6}{2}\\x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\Rightarrow z=11-0-3=8\)

Vậy nghiệm của hệ phương trình là \(\left(x;y;z\right)=\left(0;3;8\right)\).

\(\hept{\begin{cases}xy+3=3x+y\\x^2+2y^2+y=1\end{cases}}\)

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

Xét: x=1

\(\Rightarrow\left(2\right)\Leftrightarrow2y^2+y=0\Leftrightarrow\hept{\begin{cases}y=0\\y=-\frac{1}{2}\end{cases}}\)

Xét: y=3

\(\Rightarrow\left(2\right)\Leftrightarrow x^2+2.3^2+3>0\)=> vô nghiệm.

KL:.....

a: 3(x+7)-2x+5>0

=>3x+21-2x+5>0

=>x+26>0

=>x>-26

Sửa đề: \(\dfrac{x+2}{18}-\dfrac{x+3}{8}< \dfrac{x-1}{9}-\dfrac{x-4}{24}\)

=>\(\dfrac{4\left(x+2\right)}{72}-\dfrac{9\left(x+3\right)}{72}< \dfrac{8\left(x-1\right)}{72}< \dfrac{3\left(x-4\right)}{72}\)

=>\(4\left(x+2\right)-9\left(x+3\right)< 8\left(x-1\right)-3\left(x-4\right)\)

=>\(4x+8-9x-27< 8x-8-3x+12\)

=>-5x-19<5x+4

=>-10x<23

=>\(x>-\dfrac{23}{10}\)

b: \(3x+2+\left|x+5\right|=0\left(1\right)\)

TH1: x>=-5

(1) trở thành: 3x+2+x+5=0

=>4x+7=0

=>\(x=-\dfrac{7}{4}\left(nhận\right)\)

TH2: x<-5

=>x+5<0

=>|x+5|=-x-5

Phương trình (1) sẽ trở thành:

\(3x+2-x-5=0\)

=>2x-3=0

=>2x=3

=>\(x=\dfrac{3}{2}\)

\(\dfrac{3x+4}{7}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow\dfrac{6x+8}{14}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow6x+8\le5x-19\)

\(\Leftrightarrow6x-5x\le-19-8\)

\(\Leftrightarrow x\le-27\)

\(\dfrac{3x+4}{7}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow\dfrac{2\left(3x+4\right)}{14}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow2\left(3x+4\right)\le5x-19\)

\(\Leftrightarrow6x+8\le5x-19\)

\(\Leftrightarrow x\le-27\)

Vậy \(S=\left\{x|x\le-27\right\}\)

Lời giải:
Đặt $\frac{x-1}{x+2y}=a; \frac{y+1}{x-2y}=b$ thì HPT trở thành:
\(\left\{\begin{matrix} 5a+3b=8\\ 20a-7b=-6\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 20a+12b=32\\ 20a-7b=-6\end{matrix}\right.\)

\(\Rightarrow 19b=38\Rightarrow b=2\Rightarrow a=0,4\)

Ta có:

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

\(\Rightarrow 2(4y+5)-3(1+5y)=0\Rightarrow y=1\)

Kéo theo $x=3$

Vậy $(x,y)=(3,1)$

, xy*(x+y)-2x-2y tại x+y=10

->10xy-2(x+y)=10xy-20=120-20=80

b, x^5(x+2y)-x^3y*(x+2y)+x^2y^2*x+2y=(x+2y)(x^5-x^3y+x^2y^2)

Bạn tự thay vảo nhá

Vg ạ. Mình cảm ơn nhiều

\(a,\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\ b,\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\23y=46\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ d,\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

\(e,\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a. \(\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=10\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x=20\\6x-2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y=46\\5x+2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4x+3y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

e. \(\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\4x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)