bài 1: Cho tỉ lệ thức:
\(\frac{3x-y}{x+y}=\frac{3}{4}.\) Tính \(\frac{x}{y4}\)
Bài 2: Cho dãy TSBN:
\(\frac{a}{3}=\frac{b}{4}=\frac{c}{11}\). Tính \(\frac{b+c-a}{a+c-b}\)
Bài 3: Cho \(\frac{x}{3}=\frac{y}{4}\) và \(\frac{y}{5}=\frac{z}{6}\)
Tính M= \(\frac{2x+3y+4z}{3x+4y+5z}\)
Bài 4: Cho a, b, c là các số hữu tỉ \(\ne\) 0 sao cho:
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{a+b+c}{a}\)
Tính M= \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}\)
Giúp mình với, mai mình học rùi 
!!!!! Thanks nhiều !!!!!!!!!!
1)Ta có:\(\frac{3x-y}{x+y}=\frac{3}{4}\Rightarrow\left(3x-y\right)4=3\left(x+y\right)\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-3x=3y+4y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)
\(\Rightarrow\frac{x}{y4}=\frac{7}{36}\)
2)Ta có:\(\frac{a}{3}=\frac{b}{4}=\frac{c}{11}=\frac{b+c-a}{4+11-3}=\frac{b+c-a}{12}\)
\(\frac{a}{3}=\frac{b}{4}=\frac{c}{11}=\frac{a+c-b}{3+11-4}=\frac{a+c-b}{10}\)
\(\Rightarrow\frac{b+c-a}{12}=\frac{a+c-b}{10}=\frac{b+c-a}{a+c-b}=\frac{12}{10}=\frac{6}{5}\)
3)Ta có:\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{1}{5}\cdot\frac{x}{3}=\frac{y}{4}\cdot\frac{1}{5}\Rightarrow\frac{x}{15}=\frac{y}{20}\)(1)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{1}{4}\cdot\frac{y}{5}=\frac{z}{6}\cdot\frac{1}{4}\Rightarrow\frac{y}{20}=\frac{z}{24}\)(2)
Từ (1) và (2)\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=\frac{2x+3y+4z}{2\cdot15+20\cdot3+24\cdot4}=\frac{2x+3y+4z}{186}\)
=\(\frac{3x+4y+5z}{15\cdot3+20\cdot4+24\cdot5}=\frac{3x+4y+5z}{245}\)
\(\Rightarrow\frac{2x+3y+4z}{3x+4y+5z}=\frac{245}{186}\)
bài 1: \(\frac{3x-y}{x+y}=\frac{3}{4}\)hay 4(3x-y)=3(x+y)
=>12x-4y=3x+3y
12x-3x=4y+3y
9x =7y
hay \(\frac{x}{7}=\frac{y}{9}\)
Đặt \(\frac{x}{7}=\frac{y}{9}=k\) <=>\(\begin{cases}x=7k\\y=9k\end{cases}\)
=>\(\frac{x}{y4}=\frac{7k}{9k.4}=\frac{7k}{36k}=\frac{7}{36}\)
