\(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)

\(\Rightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)

\(6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd\)

\(39bc+14bc=39ad+14ad\)

\(53bc=53ad\)

\(bc=ad\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)

\(\frac{3x-7}{x+y}=\frac{3}{4}\)

\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(12x-4y=3x+3y\)

\(9x=7y\)

\(\frac{x}{y}=\frac{7}{9}\)

\(\frac{3x-y}{x+y}=\frac{3}{4}\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(\Leftrightarrow12x-4y=3x+3y\Leftrightarrow9x-7y=0\)

\(\Leftrightarrow9x=7y\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)

Vậy \(\frac{x}{y}=\frac{7}{9}\)

Bài làm

Đặt \(\frac{x}{-4}=\frac{y}{-7}=\frac{z}{3}=k\Rightarrow\hept{\begin{cases}x=-4k\\y=-7k\\z=3k\end{cases}}\)

=> \(A=\frac{-2x+y+5z}{2x-3y-6z}=\frac{-2\cdot\left(-4k\right)-7k+5\cdot3k}{2\cdot\left(-4k\right)-3\cdot\left(-7k\right)-6\cdot3k}=\frac{8k-7k+15k}{-8k+21k-18k}=\frac{16k}{-5k}=-\frac{16}{5}\)

Đặt \(x=-4k;y=-7k;z=3k\)

\(A=\frac{-2x+y+5z}{2x-3y-6z}=\frac{-2\left(-4k\right)+\left(-7k\right)+5.3k}{2\left(-4k\right)-3\left(-7k\right)-6.3k}\)

\(=\frac{8k-7k+15k}{-8k+21k-18k}=\frac{16k}{-5k}==-\frac{16}{5}\)

a, Ta có : \(Q=mn^2+n^2\left(n^2-m\right)+m^2n^4+2n^4+m^2-2\)

\(=mn^2+n^4-mn^2+m^2n^4+2n^4+m^2-2\)

\(=3n^4+m^2+m^2n^4-2\)

Đề sai ko, xem lại hộ mk nhé ! 

đề có  dấu '' /''  mà t quên , nhưng dù gì cảm ơn , t làm ra rồi