\(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=\dfrac{2x+6y-1}{5x}\left(1\right)\)

Từ `2` tỉ số đầu , ta áp dụng t/c của DTSBN , ta đc :

\(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=\dfrac{2x+3+3y-2}{3+6}=\dfrac{2x+3y+1}{9}\left(2\right)\)

Từ `(1);(2)=>`\(\dfrac{2x+6y-1}{5x}=\dfrac{2x+3y+1}{9}\left(3\right)\)

Từ `(3)` ta xét `2` trường hợp :

+, Nếu `2x+3y+1 \ne  0` thì :

`(3)=>5x=9=>x=9/5`

Thay `x=9/5` vào \(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}\), ta đc :

\(\dfrac{2\cdot\dfrac{9}{5}+3}{3}=\dfrac{3y-2}{6}\\ \Rightarrow\dfrac{\dfrac{18}{5}+3}{3}=\dfrac{3y-2}{6}\\ \Rightarrow\dfrac{11}{5}=\dfrac{3y-2}{6}\\ 3y-2=6\cdot\dfrac{11}{5}\\ 3y-2=\dfrac{66}{5}\\ 3y=\dfrac{76}{5}\\ y=\dfrac{76}{16}\)

+, Nếu `2x+3y+1=0` thì :

`(1)=>` \(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=0\\ \Rightarrow\left\{{}\begin{matrix}2x+3=0\\3y-2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)

2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)

⇒ \(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{28}\)

⇒\(9+3x=28\)

⇒\(3x=19\)

⇒\(x=\dfrac{19}{3}\)

bạn thay vào là tìm được y

Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x

=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x

=> 2x+3y-1 / 12 = 2x+3y-1 / 6x

=> 12 = 6x => x =2

a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5a+b=5\\15a=24\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{5}\\b=-3\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)

\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)

\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)

=x+y-z

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

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

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

\(a,\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2.\left(x-1\right)}+\dfrac{-2x}{\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)}{2.\left(x-1\right).\left(x+1\right)}+\dfrac{\left(-2x\right).x}{x.\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)-2x^2}{x.\left(x+1\right)\left(x-1\right)}\)

b: \(=\dfrac{y^2-12y+24}{6y\left(y-6\right)}\)

c: \(=\dfrac{12-2x+3x}{2x\left(x+3\right)}=\dfrac{x+12}{2x\left(x+3\right)}\)

4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)

Do đó: x=-16; y=-24; z=-30