\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)

thử suy nghĩ nào

a. neither do

either

b. too

c. , too

d. So was

e. too

Chắc vậy

a)either do/too

b)too

d)so was

e)too

Đặt \(\frac{3\left|x\right|+5}{3}=\frac{3\left|y\right|-1}{5}=\frac{3-z}{7}=k\)

\(\Rightarrow\left|x\right|=\frac{3k-5}{3}\Rightarrow2\left|x\right|=\frac{6k-10}{3}\)

\(\Rightarrow\left|y\right|=\frac{5k+1}{3}\Rightarrow7\left|y\right|=\frac{35k+7}{3}\)

\(\Rightarrow z=3-7k\Rightarrow3z=9-21k\)

Vì \(2\left|x\right|+7\left|y\right|+3z=-14\)\(\Rightarrow\frac{6k-10}{3}+\frac{35k+7}{3}+\left(9-21k\right)=-14\)

\(\Rightarrow\frac{\left(6k-10\right)+\left(35k+7\right)+\left(27-63k\right)}{3}=-14\)

\(\Rightarrow\frac{-22k+24}{3}=-14\)

\(\Rightarrow-22k+24=-42\)

\(\Rightarrow k=\frac{-42-24}{22}=3\)

\(\Rightarrow\left|x\right|=\frac{3.3-5}{3}=\frac{4}{3}\Rightarrow x=-\frac{4}{3};\frac{4}{3}\)

\(\Rightarrow\left|y\right|=\frac{5.3+1}{3}=\frac{16}{3}\Rightarrow y=-\frac{16}{3};\frac{16}{3}\)

\(\Rightarrow z=3-7.3=-18\)

\(\hept{\begin{cases}\left(x-y\right)⋮17\Rightarrow\left(x-y\right)=17.p...voi...P\in Z\\A-B=x^2y-xy^2=xy\left(x-y\right)=17.p.\left(xy\right)⋮17\Rightarrow dccm\Leftrightarrow dpcm\end{cases}}\)

Bạn kia ngu quá !!!!

mình giải đúng nèk

\(C=\left(a+b\right)\left(a+1\right)\left(b+1\right)=\left(a+b\right)\left[a\left(b+1\right)+\left(b+1\right)\right]\)

\(=\left(a+b\right)\left(ab+a+b+1\right)=3\left(-5+3+1\right)=3.\left(-1\right)=-3\)

\(C=\left(a+b\right)\left(a+1\right)\left(b+1\right)\)

\(C=\left(a+b\right)\cdot ab+b+a+1\)

\(C=\left(a+b\right)\cdot ab+\left(a+b\right)+1\)

Thay \(a+b=3;ab=5\)vào biểu thức \(C\)ta được \(:\)

\(C=3\cdot\left(-5\right)+3+1=-15+3+1=-11\)

Vậy \(.............................................................\)