x-y=7

nên x=y+7

\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y-7}{2y+x}\)

\(=\dfrac{3\left(y+7\right)-7}{2\cdot\left(y+7\right)+y}-\dfrac{3y-7}{2y+y+7}\)

\(=\dfrac{3y+21-7}{2y+14+y}-\dfrac{3y-7}{3y+7}\)

\(=\dfrac{3y+14}{3y+14}-\dfrac{3y-7}{3y+7}\)

\(=1-\dfrac{3y-7}{3y+7}=\dfrac{3y+7-3y+7}{3y+7}=\dfrac{14}{3y+7}\)

B=\(\frac{3x-7}{2x+y}-\frac{3y+7}{2y+x}\)

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

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

=\(\frac{2x+y}{2x+y}-\frac{2x+x}{2x+x}\)

=1-1

=0

\(\frac{4m-2n}{4m+5n}\) với \(\frac{m}{n}=\frac{1}{5}\)

Ta có : \(\frac{m}{n}=\frac{1}{5}\)hay \(\frac{m}{1}=\frac{n}{5}\)

Đặt \(\frac{m}{1}=\frac{n}{5}=k\Rightarrow\hept{\begin{cases}m=k\\n=5k\end{cases}}\)

Do đó \(\frac{4m-2n}{4m+5n}=\frac{4k-2\cdot5k}{4k+5\cdot5k}=\frac{4k-10k}{4k+25k}=\frac{-6k}{29k}=-\frac{6}{29}\)

b. \(\frac{2x+7}{3x-y}+\frac{2y-7}{3y-x}\)

Ta có : x - y = 7 => x = 7 + y

Do đó \(\frac{2x+7}{3x-y}+\frac{2y-7}{3y-x}=\frac{2\left(7+y\right)+7}{3\left(7+y\right)-y}+\frac{2y-7}{3y-\left(7+y\right)}\)

\(=\frac{14+2y+7}{21+3y-y}+\frac{2y-7}{3y-7-y}\)

\(=\frac{21+2y}{21+2y}+\frac{2y-7}{2y-7}=1+1=2\)

a) \(\frac{m}{n}=\frac{1}{5}\Rightarrow\frac{m}{1}=\frac{n}{5}\)

Đặt \(\frac{m}{1}=\frac{n}{5}=k\Rightarrow\hept{\begin{cases}m=k\\n=5k\end{cases}}\)

Thế vào ta được :

\(\frac{4m-2n}{4m+5n}=\frac{4k-2.5k}{4k+5.5k}=\frac{4k-10k}{4k+25k}=\frac{-6k}{29k}=-\frac{6}{29}\)

b) x - y = 7 => x = 7 + y

Thế vào ta được :

\(\frac{2x+7}{3x-y}+\frac{2y-7}{3y-x}=\frac{2\left(7+y\right)+7}{3\left(7+y\right)-y}+\frac{2y-7}{3y-\left(7+y\right)}\)

\(=\frac{21+2y}{21+2y}+\frac{2y-7}{3y-7-y}\)

\(=\frac{21+2y}{21+2y}+\frac{2y-7}{2y-7}=1+1=2\)

k mần đc mu

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²