Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

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

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

f: Ta có: \(x\left(2x-9\right)-4x+18=0\)

\(\Leftrightarrow\left(2x-9\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=2\end{matrix}\right.\)

g: Ta có: \(4x\left(x-1000\right)-x+1000=0\)

\(\Leftrightarrow\left(x-1000\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1000\\x=\dfrac{1}{4}\end{matrix}\right.\)

f. x(2x - 9) - 4x + 18 = 0

<=> x(2x - 9) - 2(2x - 9) = 0

<=> (x - 2)(2x - 9) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\2x-9=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{9}{2}\end{matrix}\right.\)

g. 4x(x - 1000) - x + 1000 = 0

<=> 4x(x - 1000) - (x - 1000) = 0

<=> (4x - 1)(x - 1000) = 0

<=> \(\left[{}\begin{matrix}4x-1=0\\x-1000=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=1000\end{matrix}\right.\)

h. 2x(x - 4) - 6x²(-x + 4) = 0

<=> 2x(x - 4) + 6x²(x - 4) = 0

<=> (2x + 6x²)(x - 4) = 0

<=> 2x(1 + 3x)(x - 4) = 0

<=> \(\left[{}\begin{matrix}2x=0\\1+3x=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\\x=4\end{matrix}\right.\)

i. 2x(x - 3) + x² - 9 = 0

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

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

<=> (3x + 3)(x + 3) = 0

<=> \(\left[{}\begin{matrix}3x+3=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

j. 9x - 6x² + x³ = 0

<=> x(9 - 6x + x²) = 0

<=> x(3 - x)² = 0

<=> \(\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

 4x(x – 1000) – x + 1000 = 0

(4x-1)(x-1000)                 =0

⇔x=1/4 hoặc  1000

2:

a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)

b: \(2\left(x-1\right)+x^2-x\)

\(=2\left(x-1\right)+x\left(x-1\right)\)

\(=\left(x-1\right)\left(x+2\right)\)

c: \(3x^2+14x-5\)

\(=3x^2+15x-x-5\)

\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)

3: 

a: \(2x\left(x-1\right)-2x^2=4\)

=>\(2x^2-2x-2x^2=4\)

=>-2x=4

=>x=-2

b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)

=>\(x^2-3x-\left(x^2+x-2\right)=5\)

=>\(x^2-3x-x^2-x+2=5\)

=>-4x=3

=>x=-3/4

c: \(4x^2-25+\left(2x+5\right)^2=0\)

=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)

=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)

=>4x(2x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

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

\(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(4x+2-8x-16=9\)

\(4x-8x=9+16-2\)

\(-4x=23\)

\(x=-\frac{23}{4}\)

a, \(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(\Leftrightarrow4x+2-8x-16=0\Leftrightarrow-4x-14=0\Leftrightarrow x=-\frac{7}{2}\)

b, \(\left(x+1\right)3-2x\left(x+3\right)=2\)

\(\Leftrightarrow3x+3-2x^2-6x=2\Leftrightarrow-3x+1-2x^2=0\)

a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)

\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)

\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)

\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)

\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)

Do \(\left(x+1\right)^2+1>0\)

\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

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

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

x + x(x + 3) = 0

x(1 + x + 3) = 0

x(x + 4) = 0

x = 0 hoặc x + 4 = 0

*) x + 4 = 0

x = -4

Vậy x = -4; x = 0

b) (x + 2)(x - 2) - (x + 1)² = 7

x² - 4 - x² - 2x - 1 = 7

-2x - 5 = 7

-2x = 7 + 5

-2x = 12

x = 12 : (-2)

x = -6

c) 6x² - (2x + 1)(3x - 2) = 1

6x² - 6x² + 4x - 3x + 2 = 1

x + 2 = 1

x = 1 - 2

x = -1

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

x² + 3x + 2x + 6 - x² - x + 2x + 2 = 2

6x + 8 = 2

6x = 2 - 8

6x = -6

x = -6 : 6

x = -1

e) 6(x - 1)(x + 1) - (2x - 1)(3x + 2) + 3 = 0

6x² - 6 - 6x² - 4x + 3x + 2 + 3 = 0

-x - 1 = 0

x = -1

A) -2x(3x+2)(3x-2)+5(x+2)² - (x-1)(2x+1)(2x+1)

= -2x(9x²-4)+5(x²+4x+4) - (x-1)(4x²-1)

= -18x³+8x+5x²+20x+20-(4x³-x-4x²+1)

= -18x³+5x²+28x+20-4x³+x+4x²+1

= -22x³+9x²+29x+21

B) (7x-8)(7x+8)-10(2x+3)²+5x(3x-2)²-4x(x-5)²

= 49x² - 64 -10(4x²+ 12x + 3) + 5x(9x² - 12x +4) - 4x(x² - 10x +25)

= 49x² - 64 -40x² - 120x - 30 + 45x³ - 60x² - 20x - 4x³ + 40x² -100x

= 41x³ -11x² -240x -94

C) \(\left(x^2-3\right)\left(x^2+3\right)-5x^2\left(x+1\right)^2-\left(x^2-3x\right)\left(x^2-2x\right)+4x\left(x+2\right)^2\)

\(\left(x^4-9\right)-5x^2\left(x^2+2x+1\right)-\left(x^4-2x^3-3x^3+6x^2\right)+4x\left(x^2+4x+4\right)\)

\(x^4-9-5x^4-10x^3-5x^2-x^4+5x^3-6x^2+4x^3+16x^2+16x\)

\(-5x^4-x^3+5x^2+20x-9\)

D) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)

\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)

\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)

\(-40x^4+36x^3+82x^2+6x-11\)