Ta có: a‐b = 2﴾a+b﴿

=>a‐b=2a+2b

=>‐3b=a

=>a/b=‐3/b=‐3

=>a‐b=‐3

=>‐3b‐b=‐3

=>‐4b=‐3

=>b=3/4

Mà ‐3b=a

=>a=‐3.3/4 

b= a:b ﴾b khác 0﴿

=>a=‐9/4

Thử lại: ‐9/4‐3/4=‐3

2﴾‐9/4 + 3/4﴿=‐3 ‐9/4 : 3/4 =‐3

Vậy a=‐9/4 b=3/4

thanks bn nha!!!!!!!

Ta có a:b=3:5=> \(\frac{a}{3}=\frac{b}{5}\left(1\right)\)

Lại có b:c=1:2=>\(\frac{b}{1}=\frac{c}{2}=>\frac{b}{5}=\frac{c}{10}\left(2\right)\)

Từ (1) và (2) => \(\frac{a}{3}=\frac{b}{5}=\frac{c}{10}=\frac{a+b+c}{3+5+10}=\frac{180}{18}=10\)

=> \(\hept{\begin{cases}a=30\\b=50\\c=100\end{cases}}\)

a : b = 3 : 5 

=> a = \(\frac{3}{5}b\)

b : c = 1 ; 2 

=> c = \(2b\)

Mà a + b + c = 180

=> \(\frac{3}{5}b\)+ b + 2b = 180

=> \(\left(\frac{3}{5}+1+2\right)b=180\)

=> \(\frac{18}{5}b=180\)

=> \(b=50\)

=> a = \(\frac{3}{5}b=\frac{3}{5}.50=30\)

     c = 2b = 2 . 50 = 100

Vậy a = 30 ; b = 50 ; c = 100

\(A=\)\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)

\(A=\left(\frac{9}{1}-8\right)+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)\)

\(A=1+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)

\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)

\(A=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)>\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}=B\)

\(\Rightarrow A:B=\frac{10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)\(=10\)

Vậy \(A:B=10\)

Ta có:

$a \div b = -3 \div 2 = \dfrac{a}{b} = \dfrac{-3}{2} => \dfrac{a}{-3} = \dfrac{b}{2}$

Áp dụng tc dãy tỉ số bằng nhau ta có:

$\dfrac{a}{-3} = \dfrac{b}{2} = \dfrac{a - b}{-3-2} = \dfrac{-10}{-5} =2$

`=> a/-3=b/2=2`

`=> a=-6; b=4`

\(a-b=10\)

\(\dfrac{a}{b}=\dfrac{-3}{2}\Rightarrow\dfrac{a}{-3}=\dfrac{b}{2}=\dfrac{a-b}{-3-2}=\dfrac{10}{-5}=-2\)

\(\Rightarrow a=\left(-3\right).\left(-2\right)=6\)

\(\Rightarrow b=6-10=-4\)

Vậy \(\left(a;b\right)\in\left\{6;-4\right\}\)

Đính chính

\(a-b=-10\)

...\(\dfrac{a}{-3}=\dfrac{b}{2}=\dfrac{a-b}{-3-2}=\dfrac{-10}{-5}=2\Rightarrow a=\left(-3\right).2=-6\Rightarrow b=-6+10=4\)

Vậy \(\left(a;b\right)\in\left\{-6;4\right\}\)