x⁴ - 4x³ - 8x² + 8x 

 = x(x³ - 4x² - 8x + 8) 

= x[x³ + 8 - 4x(x + 2)] 

= x[(x + 2)(x² - 2x + 4) - 4x(x + 2)] 

= x(x + 2)(x² - 6x + 4)

= x(x + 2)(x² - 6x + 9 - 5) 

 = \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)

\(x^4-4x^3-8x^2+8x\)

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

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

\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)

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

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

\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)

\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

x⁴+8x³+15x²-4x-2

= (x⁴+4x³+x²)+(4x³+16x²+4x)-(2x²+8x+2)

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

= (x²+4x+1).(x²+4x-2)

ai h minh minh h lai cho

Lời giải:

a. 

$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

a 4x -4y +(x-y)^2

=4(x-y)+(x-y).(x-y)

=(x-y).(4+x-y)

c x^2(x+1)-4(x+1)

(x+1).(x^2-4)

d x^4-(x^2-2x+1)

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

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

MIK KO BIT DUNG HAY KO CON B THI MIK KO BIET LAM

Câu b dễ thôi

\(x^4-4x^3-8x^2+8x\)

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

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

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

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

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

c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)

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

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

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

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

a) Ta có: \(8x+4x^2-12xy\)

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

b) Ta có: \(5x^3-10x^2+5x\)

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

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

c) Ta có: \(x^3+x^2y-xy^2-y^3\)

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

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

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

d) Ta có: \(x^2-8x-9\)

\(=x^2-9x+x-9\)

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

a. `8x+4x^2-12xy=4x(2+x-3y)`

b) `5x^3-10x^2+5x=5x(x^2-2x+1)`

c) `x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y)=(x+y)(x^2-y^2)=(x+y)^2 (x-y)`

d) `x^2-8x-9=(x^2-2.x.4+4^2)-25=(x-4)^2-5^2=(x+1)(x-9)`

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