Bài 13 :

Câu a : Ta có :

\(\left(3x+2\right)^2-49\)

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

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

\(=\left(3x-5\right)\left(3x+9\right)\)

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

Vì 3 chia hết cho 3 nên \(3\left(3x-5\right)\left(x+3\right)\) chia hết cho 3 .

\(\Rightarrow\left(3x+2\right)^2-49\) chia hết cho 3 ( đpcm )

Câu b : Ta có :

\(x\left(4x-1\right)^2-81x\)

\(=x\left[\left(4x-1\right)^2-9^2\right]\)

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

\(=x\left(4x-10\right)\left(4x+8\right)\)

\(=8x\left(2x-5\right)\left(x+2\right)\)

Vì 8 chia hết cho 8 nên \(8x\left(2x-5\right)\left(x+2\right)\) chia hết cho 8

\(\Rightarrow x\left(4x-1\right)^2-81x\) chia hết cho 8 ( đpcm )

Bài 14 :

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

Câu b : \(x^2+x+6\) ( Không phân tích được )

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

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

Câu e : \(x^2+4x+3=\left(x+1\right)\left(x+3\right)\)

Câu f : \(x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

Dư \(-15x^2+26x-7\)bạn nhé!

roois vãi

-Đăng tách câu hỏi bạn nhé.

rối thế bn

1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)

\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)

\(=\dfrac{1}{2}x^3+x^2-15x-18\)

2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)

\(=4x^3+6x^2-6x^2-9x+10x+15\)

\(=4x^3+x+15\)

3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)

\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)

\(=3x^5-x^4+5x^3+10x^2+26x-5\)

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

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

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

A = 6x⁴ - 5x³ + 4x² + 2x - 1

   = 6x⁴ + 3x³ - 8x³ - 4x² + 8x² + 4x - 2x - 1

   = 3x³. ( 2x + 1 ) - 4x² ( 2x + 1 ) + 4x ( 2x + 1 ) - ( 2x + 1 )

   = ( 2x + 1 ) ( 3x³ - 4x² + 4x - 1 )

    = ( 2x + 1 ) ( 3x³ - x² - 3x² + x + 3x - 1 )

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

     = ( 2x + 1 ) ( 3x - 1 ) ( x² - x + 1 )

B = 4x⁴ + 4x³ + 5x² + 8x - 6

    = 4x⁴ - 2x³ + 6x³ - 3x² + 8x² - 4x + 12x - 6

     = 2x³ ( 2x - 1 ) + 3x² ( 2x - 1 ) + 4x ( 2x - 1 ) + 6 ( 2x - 1 )

     = ( 2x - 1 ) ( 2x³ + 3x² + 4x + 6 )

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

      = ( 2x - 1 ) ( 2x + 3 ) ( x² + 2 )

C = x⁴ + x³ - 5x² + x - 6

   = x⁴ - 2x³ + 3x³ - 6x² + x² - 2x + 3x - 6 

   = x³ ( x - 2 ) + 3x² ( x - 2 ) + x ( x - 2 ) + 3 ( x - 2 )

   = ( x - 2 ) ( x³ + 3x² + x + 3 )

    = ( x - 2 ) [ x² ( x + 3 ) + ( x + 3 ) ]

    = ( x - 2 ) ( x + 3 ) ( x² + 1 ) 

\(a,x\left(x+9\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Rightarrow x\left(x^2+4x+4\right)=0\\ \Rightarrow x\left(x+2\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(x-5-4\right)\left(x-5+4\right)=0\\ \Rightarrow\left(x-9\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\\ d,\Rightarrow3\left(x+2\right)-x\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ e,\Rightarrow x^3+6x^2+12x+8-x^3-6x^2=4\\ \Rightarrow12x=-4\Rightarrow x=-\dfrac{1}{3}\\ g,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)