![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{x}{2x+6}+\frac{x}{2x-2}=\frac{3x+2}{\left(x+1\right)\left(x+3\right)}\) Đkxđ : \(x\ne-1;x\ne-3\)
⇌ x(x + 1) - x(x - 3) = 2(3x + 2)
⇌ x2 + x - x2 - 3x = 6x + 4
⇌ -8x = 4
⇌ x = \(-\frac{1}{2}\) ( tm đk)
→ S = \(\left\{-\frac{1}{2}\right\}\)
b, \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{2}{3}\) Đkxđ : \(x\ne-7\)
⇌ 30 + 24 = 2(x + 7)
⇌ 2x = 40
⇌ x = 20 (tmđk)
→ S = \(\left\{20\right\}\)
c, \(\frac{x-1}{\frac{x-1}{x+1}}=\frac{2x-1}{x^2+x}\) Đkxđ : \(x\ne-1\)
⇌ x = 2x - 1
⇌ x = 1 (tmđk)
→ S = \(\left\{1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>\(\left(\dfrac{2x+1}{9}+1\right)+\left(\dfrac{2x+2}{8}+1\right)+...+\left(\dfrac{2x+9}{1}+1\right)=0\)
=>2x+10=0
=>x=-5
b: \(\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+...+\left(\dfrac{x-2014}{2}-1\right)+\left(x-2016\right)=0\)
=>x-2016=0
=>x=2016
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(-2x+1\right)\left(x+3\right)+\left(x+1\right)\left(2x-1\right)=14\)
\(\Leftrightarrow-2x^2-6x+x+3+2x^2-x+2x-1=14\)
\(\Leftrightarrow-4x=12\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10
\(\left(2x+1\right)^2-\left(2x-3\right)\left(2x+3\right)=14\\ \Leftrightarrow4x^2+4x+1-4x^2+9=14\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)