Bạn xem lại đề nhé.

Sửa đề: \(8x^2-16x+8-32y^2\)

\(=8\left(x^2-2x+1-4y^2\right)\)

\(=8\left[\left(x^2-2x+1\right)-4y^2\right]\)

\(=8\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)

\(=8\left(x-1-2y\right)\left(x-1+2y\right)\)

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

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

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

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

\(x^4-4x^3+8x^2-16x+16 \)

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

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

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

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

a,x⁴-4x³+8x²-16x+16

=x⁴-4x³+4x²+4x²-16x+16

=x².(x-2)²+4.(x-2)²

=(x-2)²(x²+4)

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?

x^4 - 4x^3 - 8x^2 - 16x + 16 
= x^4-8x^2+16-4x^3-16x 
= ( x^2+4)^2 - 4x(x^2+4 ) 
= ( x^2 + 4 )(x^2 + 4 - 4x) 
= (x^2 + 4 )( x - 2 )^2

\(x^4-4x^3+8x^2-16x+16\)

=\(x^4-4x^3+4x^2-16x+16\)

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

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

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

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

Câu 1:

\(4x^2+16x-9\)

\(=4x^2+18x-2x-9\)

\(=2x\left(2x+9\right)-\left(2x+9\right)\)

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

Câu 2:

\(6x^2-11x+3=0\)

\(\Leftrightarrow6x^2-2x-9x+3=0\)

\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)