làm câu 2 đi. Tớ nghĩ mãi không ra

1. Ta có:\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+2b-3c}{2+6-12}=\frac{-20}{-4}=5\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{2}=5\\\frac{b}{3}=5\\\frac{c}{4}=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=10\\b=15\\c=20\end{cases}}\)

2. Ta có:\(\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{10}=\frac{b}{15}\)

\(\frac{b}{5}=\frac{c}{4}\Rightarrow\frac{b}{15}=\frac{c}{12}\)

\(\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{a-b+c}{10-15+12}=\frac{-49}{7}=-7\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{10}=-7\\\frac{b}{15}=-7\\\frac{c}{12}=-7\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=-70\\b=-105\\c=-84\end{cases}}\)

1. Ta có:a2 =b3 =c4 =a+2b−3c2+6−12 =−20−4 =5

2. Ta có:a2 =b3 ⇒a10 =b15 

b5 =c4 ⇒b15 =c12 

⇒a10 =b15 =c12 =a−b+c10−15+12 =−497 =−7

Vì abc=105

=> \(S=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)

\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(b+1+bc\right)}\)

\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)

\(=\frac{bc+b+1}{bc+b+1}=1\)

Vậy S=1.

a+b/2b=4a/b

(a+b)*b=8ab

ab+B^2=8ab

(b^2=7ab)

Bạn tự tìm ab tiếp đi ; mình đã cho thêm thông tin là (b^2=7ab).Thông cảm nha vì minh bận quá ko tìm ab được ^^

k mình nha

Vì \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=kd\)

\(\Rightarrow\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\)(1)

\(\Rightarrow\frac{2a+3c}{2b+3d}=\frac{2bk+3dk}{2b+3d}=\frac{k\left(2b+3d\right)}{2b+3d}=k\)(2)

\(\RightarrowĐPCM\)

Đặt  \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\), ta được \(a=2k;b=5k;c=7k\)Ta có:

\(\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{4k}{2k+10k-7k}=\frac{4k}{5k}=\frac{4}{5}\)

\(\Rightarrow A=\frac{4}{5}\)

Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\)

=> a = 2k ; b = 5k ; c = 7k . Thay vào A ta được :

\(A=\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{k\left(2-5+7\right)}{k\left(2+2.5-7\right)}=\frac{2-5+7}{2+2.5-7}=\frac{4}{5}\)