\(\frac{2x+1}{4}\)-\(\frac{y-2}{3}\)=\(\frac{1}{12}\)

=\(\frac{3.\left[2x+1\right]}{12}\)-\(\frac{4.\left[y-2\right]}{12}\)=\(\frac{1}{12}\)

=6x+3-4y-6=1

=6x-3-4y=1

=6x-4y=4

=2[3x-2y]=4

MK MỚI HỌC LỚP 8 ,CHÚA SẼ CHUYỂN HỆ PHƯƠNG TRÌNH CUỐI CÙNG ,BẠN GIẢI NỐT NHA 

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

\(< =>\left\{{}\begin{matrix}\text{6xy+4y-9x-6=6xy}\\\text{4xy+5y-20x-25=4xy}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{4y-9x=6}\\\text{5y-20x=25}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{20y-45x=30}\\\text{20y-80x=100}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{35x=-70}\\\text{4y-9x=6}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{x=-2}\\\text{4y-9.(-2)=6}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{x=-2}\\\text{y=-3}\end{matrix}\right.\)

vậy ...

Ta có: \(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6xy-9x+4y-6-6xy=0\\4xy-20x+5y-25-4xy=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9x+4y=6\\-20x+5y=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-180x+80y=120\\-180x+45y=225\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}35y=-105\\-9x+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\-9x=6-4y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9x=6-4\cdot\left(-3\right)=6+12=18\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(-2;-3)

nhâ vế 1 vs 2

nhân vế 2 vs 3 là ra thôi bn

trừ 2 vế cho nhau nữa

\(\left\{{}\begin{matrix}3x+2y=4\\2x-3y=7\end{matrix}\right.< =>\left\{{}\begin{matrix}6x+4y=8\\6x-9y=21\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}13y=-13\\3x+2y=4\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-1\\3x=4+2=6\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)

⇒ \(3x=-5\)

⇒ \(x=-\dfrac{5}{3}\)

\(a,\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\2y=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}3x+5y=11\\2x+5y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=11\\3x+5y-2x-5y=11-9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3.2+5y=11\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6+5y=11\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=5\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

4:

x+3y=4m+4 và 2x+y=3m+3

=>2x+6y=8m+8 và 2x+y=3m+3

=>5y=5m+5 và x+3y=4m+4

=>y=m+1 và x=4m+4-3m-3=m+1

x+y=4

=>m+1+m+1=4

=>2m+2=4

=>2m=2

=>m=1

3:

x+2y=3m+2 và 2x+y=3m+2

=>2x+4y=6m+4 và 2x+y=3m+2

=>3y=3m+2 và x+2y=3m+2

=>y=m+2/3 và x=3m+2-2m-4/3=m+2/3

giúp mình nhé

Lời giải:
Lấy PT(2) nhân 2 rồi cộng với PT(1) theo vế thu được:
$2(y-2x)+3x-2y=5.2+(-8)$

$\Leftrightarrow -x=2\Leftrightarrow x=-2$

$y=2x+5=2(-2)+5=1$

Vậy hpt có nghiệm $(x,y)=(-2,1)$