Ta có : \(\frac{x}{y}=\frac{3}{5}\Rightarrow x=\frac{3}{5}y\)

Thay vào biểu thức trên ta được : 

\(\frac{x+y}{x-y}=-4\Rightarrow\frac{\frac{3}{5}y+y}{\frac{3}{5}y-y}=-4\)

\(\Rightarrow\frac{\frac{8}{5}y}{-\frac{2}{5}y}=-4\Rightarrow\frac{8}{5}y=\frac{8}{5}y\)( đúng )

Vậy ta có đpcm 

んuﾘ ｲ mày không hiểu đề bài à Tú :))

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

<=> x + y = -4x + 4y 

<=> x + y + 4x - 4y = 0

<=> 5x - 3y = 0

<=> 5x = 3y

<=> x/y = 3/5 ( đpcm )

hình như em đánh thừa số 1 rồi 

\(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{707070}{151515}\)

\(=\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{93}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{707070}{151515}\)

\(=\frac{7}{15}-\frac{707070}{151515}=\frac{7}{15}-\frac{7}{15}=0\)

(do \(\frac{707070}{151515}=\frac{7.101010}{15.101010}=\frac{7}{15}\))

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)

\(=1-\frac{1}{2n+3}=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)

\(\frac{1}{30}+\frac{1}{42}+...+\frac{1}{182}+\frac{1}{210}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{13.14}+\frac{1}{14.15}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}=\frac{1}{15}\)

\(=\frac{1}{5}-\frac{1}{15}\)

\(=\frac{3}{15}-\frac{1}{15}\)

\(=\frac{2}{15}\)

\(x-\frac{2}{5}=1-\frac{x}{6}\)

\(x+\frac{x}{6}=1+\frac{2}{5}\)

\(x+\frac{x}{6}=\frac{7}{5}\)

\(\frac{30x}{30}+\frac{5x}{30}=\frac{42}{30}\)

\(35x=42\)

\(x=\frac{42}{35}\)

hellllllllllllllllllo 

= 14972

 nha

Xin lỗi viết nhầm  = -14972