\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)

#)Giải :

\(x^3-2x-4\)

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

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

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

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

\(x^4+2x^3+5x^2+4x-12\)

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

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

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

\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

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

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

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

cảm ơn nha!

      \(x^4+2x^2-24\)

\(=x^4+6x^2-4x^2-24\)

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

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

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

       \(x^3-2x-4\)

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

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

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

Chúc bạn học tốt.

\(\left(x^2\right)^2+2x^2-24\)

a) x⁴ + 2x² - 24

= x⁴ - 4x² + 6x² - 24

= x².(x²-4) + 6.(x²-4)

= (x²-4).(x² +6)

= (x+2).(x-2).(x² +6)

( x² - 3x )² + ( 2x² - 6x ) - 24

= ( x² - 3x )² + 2( x² - 3x ) - 24 (*)

Đặt t = x² - 3x 

(*) trở thành :

t² + 2t - 24

= t² - 4t + 6t - 24

= t( t - 4 ) + 6( t - 4 )

= ( t - 4 )( t + 6 )

= ( x² - 3x - 4 )( x² - 3x + 6 )

= ( x² + x - 4x - 4 )( x² - 3x + 6 )

= [ x( x + 1 ) - 4( x + 1 ) ]( x² - 3x + 6 )

= ( x + 1 )( x - 4 )( x² - 3x + 6 )

\(\left(x^2-3x\right)^2+\left(2x^2-6x\right)-24\)

\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-24\)(1)

Đặt \(a=x^2-3x\)

(1)=\(a^2+2a-24\)

\(=a^2-4a+6a-24\)

\(=a\left(a-4\right)+6\left(a-4\right)\)

\(=\left(a-4\right)\left(a+6\right)\)

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

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

\(=\left[x\left(x-4\right)+\left(x-4\right)\right]\left(x^2-3x+6\right)\)

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(\left(x+1\right)\left(x+4\right)\right)\left(\left(x+2\right)\left(x+3\right)\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

- Đặt \(x^2+5x+5=a\)

\(=\left(a-1\right)\left(a+1\right)-24=a^2-1-24=a^2-25\)

\(=\left(a-5\right)\left(a+5\right)\)

Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25\)

\(=\left(x^2+7x+11\right)^2-25=\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

=[(x+2)(x+5)][(x+3)(x+4)]−24=(x2+7x+10)(x2+7x+12)−24=(x2+7x+11)2−1−24=(x2+7x+11)2−25=(x2+7x+11−5)(x2+7x+11+5)=(x2+7x+6)(x2+7x+16)=(x+1)(x+6)(x2+7x+16)