a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)

Thay:

\(\frac{ab}{cd}=\frac{b^2}{d^2}\)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)

=> đpcm

Bài 1:

b) \(2x+6⋮x-3\)

\(\Leftrightarrow2\left(x-3\right)+12⋮x-3\)

Mà \(2\left(x-3\right)⋮x-3\)

\(\Rightarrow12⋮x-3\)

làm nốt

d) \(x-1⋮2x+1\)

\(\Leftrightarrow2x-2⋮2x+1\)

\(\Leftrightarrow2x+1-3⋮2x+1\)

Mà \(2x+1⋮2x+1\)

\(\Rightarrow3⋮2x+1\)

Làm nốt

\(\left(x+2\right)^5=2^{10}\)

\(\left(x+2\right)^5=4^5\)

\(\Rightarrow x+2=4\)

\(x=2\)

Vậy x=2

\(\left(x+2\right)^5=2^{10}\)

\(\left(x+2\right)^5=\left(2^2\right)^5\)

\(\left(x+2\right)^5=4^5\)

\(x+2=4\)

\(x=4-2\)

\(x=2\)

=> 3x-4=0 hoặc x-1=0

=> x=4/3 hoặc x=1

\(\left(3x-4\right)\left(x-1\right)^3=0\)

\(\Rightarrow\orbr{\begin{cases}3x-4=0\\\left(x-1\right)^3=0\end{cases}}\)               \(\Rightarrow\orbr{\begin{cases}3x=0+4\\x-1=0\end{cases}}\)                 \(\Rightarrow\orbr{\begin{cases}3x=4\\x=0+1\end{cases}}\)               \(\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)

\(\Rightarrow\text{ }x\in\text{ }\left\{\frac{4}{3}\text{ ; }1\right\}\)

a) \(\left|3x-1\right|+2=x\)

=> \(\left|3x-1\right|=x-2\)(Đk: \(x-2\ge0\) <=> \(x\ge\)2)

=> \(\orbr{\begin{cases}3x-1=x-2\\3x-1=2-x\end{cases}}\)

=> \(\orbr{\begin{cases}2x=-1\\4x=3\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\) (ktm)

b) \(\left|x+15\right|+1=3x\)

=> \(\left|x+15\right|=3x-1\)(Đk: \(3x-1\ge0\) <=> \(x\ge\frac{1}{3}\))

=> \(\orbr{\begin{cases}x+15=3x-1\\x+15=1-3x\end{cases}}\)

=> \(\orbr{\begin{cases}-2x=-16\\4x=-14\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=-\frac{7}{2}\left(ktm\right)\end{cases}}\)

c. \(\left|2x-3\right|-x=2\)
\(\Leftrightarrow\left|2x-3\right|=2+x\)(1) 
Neu: \(2x-3\ge0\Leftrightarrow x\ge\frac{3}{2}\)
\(pt\left(1\right)\Leftrightarrow2x-3=2+x\)
\(\Leftrightarrow x=5\left(TM\right)\)
Neu: \(2x-3< 0\Leftrightarrow x< \frac{3}{2}\)
\(pt\left(2\right)\Leftrightarrow2x-3=-2-x\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\frac{1}{3}\left(TM\right)\)
Vay \(S=\left\{5;\frac{1}{3}\right\}\)
 

trường bạn thì mình k bt

thanks ban tra loi nha nhung ban co the cho mk biet truong ban co lich nhu the nao k ?  tai mk hoi gap 

:))

Với y = 15, ta có :

-2x = 5 . 15

-2x = 75

x = 75 : ( -2 )

x = -75/2

Vậy x = 15 ; y = -75/2

Ta có : \(-2x=5.15=75\)

\(\Rightarrow x=75:\left(-2\right)=-\frac{75}{2}\)

Vậy \(x=-\frac{75}{2};y=15\)

Để  \(A=x^2\Rightarrow x^2=x^2-3x-4\) 

\(\Rightarrow3x+4=0\)

\(\Rightarrow3x=-4\)

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

Vậy x=-4/3

x=-4/3 X=4/5