Ta có: \(\left\{{}\begin{matrix}2\left(x-y\right)+3x=1\\3x+2\left(x-y\right)=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=1\\5x-2y=7\end{matrix}\right.\)(Vô lý)

Vậy: Hệ phương trình vô nghiệm

\(hpt\text{⇔}\left\{{}\begin{matrix}2x-2y+3x=1\\3x+2x-2y=7\end{matrix}\right.\)\(\text{⇔}\left\{{}\begin{matrix}5x-2y=1\\5x-2y=7\end{matrix}\right.\)

Ta thấy : \(\dfrac{5}{5}=\dfrac{-2}{-2}\ne\dfrac{1}{7}\)

Suy ra hệ phương trình vô nghiệm

a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)

$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$

`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$

`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$

Vậy HPT vô nghiệm

$\begin{cases}|x-2|+2|y-1|=9\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|+2|y-1|=9\\2x+2|y-1|=-2\\\end{cases}$

`<=>` $\begin{cases}|x-2|-2x=11\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|=2x+11\\x+|y-1|=-1\\\end{cases}$

Đến đây dễ rồi bạn tự giải :D

1. \(\left\{{}\begin{matrix}3x^2+y^2+4xy=8\left(1\right)\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)

=> \(3x^2+3xy+xy+y^2=\left(x+y\right)\left(x^2+xy+2\right)\)

<=> \(\left(x+y\right)\left(3x+y\right)=\left(x+y\right)\left(x^2+xy+2\right)=0\)

<=> \(\left(x+y\right)\left(x^2+xy+2-3x-y\right)=0\)

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

TH1: x = -y thay vào pt (1), ta được:

3y² + y² - 4y² = 8

<=> 0y = 8 (vô lí)

TH2: \(x^2+xy+2-3x-y=0\)

<=> x (x + y) - (x + y) - 2(x - 1) = 0

<=> (x - 1)(x + y) - 2(X - 1) = 0

<=> (x - 1)(x + y - 2) = 0

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

Với x =  1 thay vào pt (1) -> 3 + y² + 4y = 8

<=> y² + 4y - 5 = 0 <=> (y + 5)(y - 1) = 0

<=> \(\left[{}\begin{matrix}y=-5\\y=1\end{matrix}\right.\)

Với x + y - 2 = 0 => x = 2 - y thay vào pt (1)

=> 3(2 - y)² + y² + 4(2 - y)y = 8

<=> 3y² - 12y + 12 + y² + 8 - 4y² = 8

<=> 12 = 12y <=> y= 1 => x = 2 - 1 = 1

Vậy ....

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

ĐKXĐ: ...

\(y\left(y^2-5y+4\right)+y^2=\left(y^2-5y+4\right)\sqrt{x+1}+x+1\)

\(\Leftrightarrow\left(y^2-5y+4\right)\left(y-\sqrt{x+1}\right)+\left(y+\sqrt{x+1}\right)\left(y-\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\left(y-\sqrt{x+1}\right)\left[\left(y-2\right)^2+\sqrt{x+1}\right]=0\)

\(\Leftrightarrow y=\sqrt{x+1}\Rightarrow y^2=x+1\)

Thế xuống pt dưới:

\(2\sqrt{x^2-3x+3}+6x-7=\left(x+1\right)\left(x-1\right)^2+x\sqrt{3x-2}\)

\(\Leftrightarrow2\left(\sqrt{x^2-3x+3}-1\right)+x\left(x-\sqrt{3x-2}\right)=x^3-7x+6\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{\sqrt{x^2-3x+3}+1}+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=\left(x+3\right)\left(x^2-3x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\\\dfrac{2}{\sqrt{x^2-3x+3}+1}+\dfrac{x}{x+\sqrt{3x-2}}=x+3\left(1\right)\end{matrix}\right.\)

Xét (1) với \(x\ge\dfrac{3}{2}\):

\(\dfrac{2}{\sqrt{x^2-3x+3}+1}\le8-4\sqrt{3}< 1\)

\(\sqrt{3x-2}\ge0\Rightarrow\dfrac{x}{x+\sqrt{3x-2}}\le1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x^2-3x+3}+1}+\dfrac{x}{x+\sqrt{3x-2}}< 2\\x+3>2\end{matrix}\right.\) 

\(\Rightarrow\left(1\right)\) vô nghiệm

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)

=>xy-2x=xy-4x+2y-8 và 2xy+7x-6y-21=2xy+6x-7y-21

=>2x-2y=-8 và x+y=0

=>x-y=-4 và x+y=0

=>2x=-4 và x+y=0

=>x=-2 và y=2