\(\Leftrightarrow x^3+2x^2-3x-x^3-3x^2=-4\)

\(\Leftrightarrow x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>x=-4 hoặc x=1

bài này mình biết làm nhưng thôi

\(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)^2=8.2\)

=>(x+1)²=16

=>x+1=\(\sqrt{16}\)

=>x+1=4

=>x=3

`x-1/8=1/4xx2/3`

`x-1/8=1/6`

`x=1/6+1/8`

`x=7/24`

x - 1/8 = 1/6

x         = 6/48 + 8/48
x         = 14/48

x - 1/8 = 1/4 x 2/3

x - 1/8=   1/6

x        = 1/6 + 1/8

x       =  14/48

Bài 2: 

Ta có: \(x\left(x-4\right)-x^2+8=0\)

\(\Leftrightarrow x^2-4x-x^2+8=0\)

\(\Leftrightarrow-4x=-8\)

hay x=2

1)=x²-49-x²

=-49

2)=>x²-4x-x²+8=0

=>-4x+8=0

=>-4x=-8

=>x=2

\(x+\left(x+1\right)+\left(x+2\right)+...+19+20=20\)

\(\Leftrightarrow x+\left(x+1\right)+\left(x+2\right)+...+19=0\)

\(\Leftrightarrow\frac{x\left(19+x\right)}{2}=0\)

\(\Leftrightarrow x\left(x+19\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+19=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-19\end{cases}}\)

Mà \(x\ne0\)nên \(x=-19\)

\(x+x.\dfrac{1}{3}+3y=\dfrac{1}{5}\)

\(x.\dfrac{4}{3}+3y=\dfrac{1}{5}\)

\(x.\dfrac{4}{3}=\dfrac{1}{5}-3y\)

\(x=\left(\dfrac{1}{5}-3y\right):\dfrac{4}{3}=\dfrac{3}{20}-\dfrac{9y}{4}\)

x : 1/3 = 1/2 +1/3

=>x:1/3=5/6

=>x=5/6 x 1/3

=>x=5/18

(x:2/3):2/5=7/10

=>x:2/3=7/10 x 2/5 

=>x:2/3=7/25

=>x=7/25x2/3

=>x=14/75

CHÚC BẠN HỌC TỐT

cái này vừa gửi rồi mà:D

x:1/3=1/2+1/3
x:1/3=5/6
x      =5/6.1/3
x      =5/18
 

(x:2/3):2/5=7/10
x:2/3        =7/10.2/5
x:2/3        =7/25
x:2/3        =7/25.2/3
x:2/3        =14/75