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![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
1) \(2x\left(x-5\right)+\left(x-2\right)\left(x+3\right)=2x^2-10x+x^2+3x-2x-6=3x^2-9x-6\)
2) \(\left(2x-5\right)\left(1-x\right)-\left(x-3\right)\left(-2x\right)=2x-2x^2-5+5x+2x^2-6x=x-5\)
3) \(\left(4x-3\right)\left(4x-3\right)-\left(3x+2\right)\left(3x-2\right)=\left(4x-3\right)^2-9x^2+4=16x^2-24x+9-9x^2+4\)
\(=7x^2-24x+13\)
4) \(\left(2x-1\right)\left(2x+1\right)\left(2x+1\right)-4\left(x^2+1\right)=\left(2x-1\right)[\left(2x+1\right)^2]-4x^2-4\)
\(=\left(2x-1\right)\left(4x^2+4x+4\right)-4x^2-4=8x^3+8x^2+8x-4x^2-4x-4-4x^2-4=8x^3+4x-8\)
5) \(3x\left(2x-8\right)-\left(2-6x\right)\left(5+x\right)=6x^2-24x-10-2x+30x+6x^2=12x^2+4x-10\)
6) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+6x+12x-24+8=-16\)
7) \(\left(x+2\right)\left(x^2-2x+4\right)-x^2\left(x-2\right)-2x^2=x^3+8-x^3+2x^2-2x^2=8\)