coi lại đề bạn nhé :D

\(x^3+2x^2-2x-12=x^3-2x^2+4x^2-8x+6x-12\)

\(=x^2\left(x-2\right)+4x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(x^2+4x+6\right)\)

\(x^3+2x^2-2x-12\)

\(=x^3-2x^2+4x^2-8x+6x-12\)

\(=x^2\left(x-2\right)+4x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+4x+6\right)\)

hk tốt

^^

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

a)\(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow2x^3+2x+3x^2+3=0\)

\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\x^2+1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=-3\\x^2+1>0\left(loai\right)\end{array}\right.\)

\(\Leftrightarrow x=-\frac{3}{2}\)

b)\(x\left(2x-1\right)\left(1-2x\right)=0\)

\(\Leftrightarrow-x\left(2x-1\right)\left(2x-1\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\\left(2x-1\right)^2=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\2x=1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)

\(2x^3+3x^2+2x+3=0\)

\(2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

\(\left(2x+3\right)\left(x^2+1\right)=0\)

\(2x+3=0\left(x^2+1\ge1>0\right)\)

\(2x=-3\)

\(x=-\frac{3}{2}\)

\(x\left(2x-1\right)\left(1-2x\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\2x-1=0\\1-2x=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\2x=1\\2x=1\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)

a)2x³+3x²+2x+3=0

(2x³+2x)+(3x²+3)=0

2x​(x²+1)+3(x²+1)=0

(x​²+1)(2x+3)=0

Vì x luôn luôn lớn hơn hoặc bằng 0 (với mọi x). Nên x+1>0 (với mọi x)

→ 2x+3=0

->x=-3/2

Vậy x=-3/2

b)x(2x-1)(1-2x)=0

=>x=0

=> 2x-1=0 => x=1/2

=>1-2x=0 => x=1/2

 Vậy x=0 hoặc x=1/2.

haha quá dễ

c, x mũ 2 + 3x - 4 = x^2 + 3 x X -4

c, x mũ 2 + 3x - 18= x^2 + 3xX -18

c, 2x mũ 2 + 3x - 5=  2xX^2 + 3xX -5

c, 3x mũ 2 - 8x + 4= 3 x X^2 - 8 x X + 4

c, 8x mũ 2 + 2x - 3= 8 x X^2 + 2 x X -3

a) = x² - 4x + x - 4 = x( x - 4 ) + ( x - 4 ) = ( x - 4 )( x + 1 )

b) = x² - 3x + 6x - 18 = x( x - 3 ) + 6( x - 3 ) = ( x + 6 )( x - 3 )

c) = 2x² - 2x + 5x - 5 = 2x( x - 1 ) + 5( x - 1 ) = ( x - 1 )( 2x + 5 )

d) = 3x² - 6x - 2x + 4 = 3x( x - 2 ) - 2( x - 2 ) = ( x - 2 )( 3x - 2 )

e) = 8x² - 4x + 6x - 3 = 4x( 2x - 1 ) + 3( 2x - 1 ) = ( 2x - 1 )( 4x + 3 )

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

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

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

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

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

a) \(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2=\left(9-3x-2\right)\left(9+3x+2\right)=\left(7-3x\right)\left(11+3x\right)\)

b) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)

c) \(9\left(x-5y\right)^2-16\left(x+y\right)^2=\left[3\left(x-5y\right)-4\left(x+y\right)\right]\left[3\left(x-5y\right)+4\left(x+y\right)\right]\)

\(=\left(-x-19y\right)\left(7x-11y\right)\)