Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

\(a,=12x^2-4x-6x-2-x-3=12x^2-11x-5\\ b,=12x^2-9x-12x^2-4x+5=5-13x\\ c,=12x^3-4x^2-12x^3-12x^2+7x-3=-16x^2+7x-3\\ d,=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)

\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)

\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)

\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)

Bài 4: 

b: Ta có: \(B=3x^2\left(x^2-5\right)+x\left(-3x^3+4x\right)+6x^2\)

\(=3x^4-15x^2-3x^3+4x^2+6x^2\)

\(=-5x^2\)

\(=-5\cdot25=-125\)

a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

= 2x² + 6x – (3x³ + 6x²) + (3x³ + 4x² - 6x)

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

= (– 3x³ + 3x³ ) + (2x²  - 6x² + 4x² ) + (6x – 6x)

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

1: =>(x+2)^2-3|x+2|=0

=>|x+2|(|x+2|-3)=0

=>x+2=0 hoặc x+2=3 hoặc x+2=-3

=>x=-2; x=1; x=-5

\(a,x-5\left(x-2\right)=6x\\ \Leftrightarrow x-5x+10-6x=0\\ \Leftrightarrow-10x+10=0\\ \Leftrightarrow x=1\\ b,2^3+3x^2-32x=48\\ \Leftrightarrow3x^2-32x+8=48\\ \Leftrightarrow3x^2-32x-40=0\)

Nghiệm xấu lắm bn

\(c,\left(3x+1\right)\left(x-3\right)^2=\left(3x+1\right)\left(2x-5\right)^2\\ \Leftrightarrow c,\left(3x+1\right)\left[\left(2x-5\right)^2-\left(x-3\right)^2\right]\\ \Leftrightarrow\left(3x+1\right)\left(2x-5-x+3\right)\left(2x-5+x-3\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x-2\right)\left(3x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)

\(d,9x^2-1=\left(3x+1\right)\left(4x+1\right)\\ \Leftrightarrow\left(3x+1\right)\left(4x+1\right)-\left(3x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(4x+1-3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)

\(b,2x^3+3x^2-32x-48=0\\ \Leftrightarrow\left(2x^3-8x^2\right)+\left(11x^2-44x\right)+\left(12x-48\right)=0\\ \Leftrightarrow2x^2\left(x-4\right)+11x\left(x-4\right)+12\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2x^2+11x+12\right)=0\\ \Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)+\left(3x+12\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)+3\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left(2x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{3}{2}\\x=-4\end{matrix}\right.\)

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

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

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