a) 3x2 – 7x + 2

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

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

\(=3x\left(x-2\right)-\left(x-2\right)\)

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

 b) a(x2 + 1) – x(a2 + 1)

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

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

.......?

a) Ta có: \(3x^2-7x+2\)

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

\(=3x\left(x-2\right)-\left(x-2\right)\)

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

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

\(=x^2a+a-a^2x-x\)

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

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

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

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

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

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

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

Phân tích đa thức thành nhân tử:

1. x² - 5x - 6

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

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

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

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

3. 1 + 3x + 2x²

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

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

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

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

4. 6 - 2x - 8x²

\(=6-8x+6x-8x^2\)

\(=\left(6+6x\right)-\left(8x+8x^2\right)\)

\(=6\left(1+x\right)-8x\left(1+x\right)\)

\(=\left(6-8x\right)\left(1+x\right)\)

5. 7 - 4x - 3x²

\(=7-7x+3x-3x^2\)

\(=\left(7-7x\right)+\left(3x-3x^2\right)\)

\(=7\left(1-x\right)+3x\left(1-x\right)\)

\(=\left(7+3x\right)\left(1-x\right)\)

1) \(x^2-5x-6\\ =x^2+x-6x-6\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x-6\right)\left(x+1\right)\)

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

1) \(x^2-8x+7=0\)

\(\Leftrightarrow x^2-7x-x+7=0\)

\(\Leftrightarrow\left(x^2-7x\right)-\left(x-7\right)=0\)

\(\Leftrightarrow x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

2) \(5x^2-11x+6=0\)

\(\Leftrightarrow5x^2-5x-6x+6=0\)

\(\Leftrightarrow\left(5x^2-5x\right)-\left(6x-6\right)=0\)

\(\Leftrightarrow5x\left(x-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=1\end{matrix}\right.\)

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

\(\Leftrightarrow2x^2-2x-x+1=0\)

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

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

4) \(x^2+7x-8=0\)

\(\Leftrightarrow x^2+8x-x-8=0\)

\(\Leftrightarrow\left(x^2+8x\right)-\left(x+8\right)=0\)

\(\Leftrightarrow x\left(x+8\right)-\left(x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\)

5) \(3x^2+7x-10=0\)

\(\Leftrightarrow3x^2-3x+10x-10=0\)

\(\Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\)

a: x^3-7x-6

=x^3-x-6x-6

=x(x-1)(x+1)-6(x+1)

=(x+1)(x^2-x-6)

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

b: =2x^3+x^2-2x^2-x+6x+3

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

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

c: =2x^3-3x^2-2x^2+3x+2x-3

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

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

d: =2x^3+x^2+2x^2+x+2x+1

=(2x+1)(x^2+x+1)

e: =3x^3+x^2-3x^2-x+6x+2

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

f: =27x^3-9x^2-18x^2+6x+12x-4

=(3x-1)(9x^2-6x+4)

a) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

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

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

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

b) \(2x^3-x^2+5x+3\)

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

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

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

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

c) \(2x^3-5x^2+5x+1\)

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

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

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

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

d) \(2x^3+3x^2+3x+1\)

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

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

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

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

e) \(3x^3-2x^2+5x+2\)

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

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

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

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

f) \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

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

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

1. \(x^2+3x+2=x^2+2x+x+2=x\left(x+2\right)+\left(x+2\right)=\left(x+2\right)\left(x+1\right)\)
2.
\(3x^2+2x-1=3x^2+3x-x-1=3x\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(3x-1\right)\)3 và 4 bạn xem lại đề nha mik giải ko đc
5. \(x^2+2x-8=x^2+4x-2x-8=x\left(x+4\right)-2\left(x+4\right)=\left(x+4\right)\left(x-2\right)\)

3: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

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

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

b) \(B=\left(x^3-3x^2\right)-\left(4x-12\right)\)

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

a) A= x^3 + x^2-x-1

= (x^3 + x^2)  - (x+1)

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

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

b)B= x^3 - 3x^2 - 4x +12

= (x^3 - 3x^2)  - (4x - 12)

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

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

c)C= -x^3 + 3x - 2

= 

d)D= 2x^3 - x^2 - 8x + 4

= (2x^3 - x^2) - (8x -4)

= x^2(2x -1) -  4(2x -1)

= (x^2 -4) (2x-1)

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

đến đây thôi

AI LÀM MÌNH SẼ K ĐÚNG + 1 THẺ KHÓA HỌC T.ANH GIAO TIẾP 

\(a.2x^2+5x+2=2x^2+4x+x+2=\left(x+2\right)\left(2x+1\right)\)

Minh Anh ơi, bạn làm hết được không?

1) 4x² + 5x - 6 = 4x² + 8x - 3x - 6 = 4x( x + 2 ) - 3( x + 2 ) = ( x + 2 )( 4x - 3 )

2) 5x² - 18x - 8 = 5x² - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )

3) 2x² + 3x - 27 = 2x² - 6x + 9x - 27 = 2x( x - 3 ) + 9( x - 3 ) = ( x - 3 )( 2x + 9 ) < đã sửa ._. >

4) 7x² + 3xy - 10y² = 7x² - 7xy + 10xy - 10y² = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )

5) x² + 5x - 2 < sai đề ._. >

6) x⁸ + x⁷ + 1 = x⁸ + x⁷ + x⁶ - x⁶ + 1

= ( x⁸ + x⁷ + x⁶ ) - ( x⁶ - 1 )

= x⁶( x² + x + 1 ) - ( x³ - 1 )( x³ + 1 )

= x⁶( x² + x + 1 ) - ( x - 1 )( x² + x + 1 )( x³ + 1 )

= ( x² + x + 1 )[ x⁶ - ( x - 1 )( x³ + 1 ) ]

= ( x² + x + 1 )( x⁶ - x⁴ + x³ - x + 1 )

1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)

\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)

\(=x^3.\left(x+2\right)+4x.\left(x+2\right)-5.\left(x+2\right)=\left(x+2\right)\left(x^3+4x-5\right)\)

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

2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

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

Đặt  \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)

\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)

Thay a , ta có :

\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)

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