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\(x^4-32x^2-16x+255=x^4+5x^3-7x^2-51x-5x^3-25x^2+35x+255\)
\(=\left(x^4+5x^3-7x^2-51x\right)-\left(5x^3+25x^2-35x-255\right)\)
\(=x\left(x^3+5x^2-7x-51\right)-5\left(x^3+5x^2-7x-51\right)\)
\(=\left(x^3+5x^2-7x-51\right)\left(x-5\right)\)
\(=\left[\left(x^3+8x^2+17x\right)-\left(3x^2-24x-51\right)\right]\left(x-5\right)\)
\(=\left[x\left(x^2+8x+17\right)-3\left(x^2+8x+17\right)\right]\left(x-5\right)\)
\(=\left(x^2+8x+17\right)\left(x-3\right)\left(x-5\right)\)

1.\(x^3+6x^2+12xy+8=x^3+3.2x^2+3.2^2x+2^3=\left(x+2\right)^3\)
3.\(x^4+2x^3+x^2-y^2=\left(x^2\right)^2+2x^2.x+x^2-y^2\)\(=\left(x^2+x\right)^2-y^2=\left(x^2+x-y\right)\left(x^2+x+y\right)\)
k mình nha bn !!!!!!! cái 2 bn xem lại đề đi, rồi mình giải cho

\(1.x^3+2x+x^2=x\left(x^2+x+2\right)\)
\(2.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
\(3.-3x^3-5x^2+8x=-3x^3+3x^2-8x^2+8x\)
\(=-3x^2\left(x-1\right)-8x\left(x-1\right)=\left(3x^2+8x\right)\left(1-x\right)\)
\(=x\left(3x+8\right)\left(1-x\right)\)
\(4.x^2+4x-5=x^2-x+5x-5=\left(x-1\right)\left(x+5\right)\)
\(5.6x^2-3x-3=6x^2-6x+3x-3=3\left(x-1\right)\left(2x+1\right)\)
\(6.3x^2-2x-5=3x^2+3x-5x-5=\left(x+1\right)\left(3x-5\right)\)
\(8.x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)\(=\left(x+2y\right)\left(x-y-2\right)\)
\(9.x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
\(10.x^2-y^2+6x+9=\left(x+3-y\right)\left(x+3+y\right)\)
Cac ban giup minh voi
1) Giai cac phuong trinh
a) 2010.(4x-3)-4x2+3=0
b)( x2-\(\frac{25}{4}\))2= 10x +1


a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....

Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)

a. x^2 + 5x + 25/4 = x^2 + 2.x .5/2 + (5/2)^2 = ( x + 5/2)^2
b, 16x^2 - 8x + 1 = (4x)^2 - 2.4x.1 + 1 = ( 4x - 1 )^2
c, 4x^2 + 12xy + 9y^2 = (2x)^2 + 2.2x.3y + (3y)^2 = ( 2x + 3y)^2

a) \(H=x^2-4x+16\)
\(H=\left(x+2\right)^2+12\ge12\)
vậy min H=12 \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Trả lời:
9y2 + 8x - 4x2 - 4
= 9y2 - ( 4x2 - 8x + 4 )
= ( 3y )2 - [ ( 2x )2 - 2.2x.2 + 22 ]
= ( 3y )2 - ( 2x - 2 )2
= [ 3y - ( 2x - 2 ) ] [ 3y + ( 2x - 2 ) ]
= ( 3y - 2x + 2 ) ( 3y + 2x - 2 )
9y2+8x-4x2-4
=9y2-(4x2-8x+4)
=9y2-(2x-2)2
=(9y-2x+2)(9x+2x-2)
#H