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a)x3-6x2+9x=x(x2-6x+9)=x(x-3)2
b)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
c)x2-x+xy-y=x(x-1)+y(x-1)=(x-1)(x+y)
d)3x2-6xy-75+3y2=3[(x2-2xy+y2)-25]=3[(x-y)2-52]=3(x-y-5)(x-y+5)
e)2x2-5x-7=(2x2+2x)-(7x+7)=2x(x+1)-7(x+1)=(x+1)(2x-7)
f)x4+36=x4+12x2+36-12x2=(x2+6)2-12x2=(x2-\(\sqrt{12}x\)+6)(x2+\(\sqrt{12}x\)+6)
h)x4+4y4=x4+4x2y2+4y2-4x2y2=(x2+2y2)-4x2y2=(x2+2y2-2xy)(x2+2y2+2xy)
a) \(4x^2-4xy+y^2-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y+3\right)\left(2x-y-3\right)\)
b) \(x^2-36+4xy+4y^2\)
\(=\left(4y^2+4xy+x^2\right)-36\)
\(=\left(2y+x\right)^2-6^2\)
\(=\left(2y+x+6\right)\left(2y+x-6\right)\)
c) \(9x^2-12xy-25+4y^2\)
\(=\left(9x^2-12xy+4y^2\right)-25\)
\(=\left(3x-2y\right)^2-5^2\)
\(=\left(3x-2y+5\right)\left(3x-2y-5\right)\)
d) \(25x^2+10x-4y^2+1\)
\(=\left(25x^2+10x+1\right)-4y^2\)
\(=\left(5x+1\right)^2-\left(2y\right)^2\)
\(=\left(5x+2y+1\right)\left(5x-2y+1\right)\)
a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))
a) x^2 + 2x + 1
=\(x^2+2.x.1+1^2\)
\(=\left(x+1\right)^2\)
b) x^2 + 6x + 9
=\(x^2+2.x.3+3^2\)
\(=\left(x+3\right)^2\)
c) x^2 - 6x + 9
\(=x^2-2.x.3+3^2\)
=\(\left(x-3\right)^2\)
d) x^2 - 2x + 1
\(=x^2-2.x.1+1^2\)
\(=\left(x-1\right)^2\)
e ) 4x^2 + 4xy +y^2
\(=\left(2x\right)^2+2.2x.y+y^2\)
\(=\left(2x+y\right)^2\)
f) x^2 + 4xy + 4y^2
\(=x^2+2.x.2y+\left(2y\right)^2\)
\(=\left(x+2y\right)^2\)
phân tích các đa thức sau thành nhân tử
a) x^2 + 2x + 1
\(=\left(x+1\right)^2\)
b) x^2 + 6x + 9
\(=\left(x+3\right)^2\)
c) x^2 - 6x + 9
\(=\left(x-3\right)^2\)
d) x^2 - 2x + 1
\(=\left(x-1\right)^2\)
e ) 4x^2 + 4xy +y^2
\(=\left(2x\right)^2+4xy+y^2\)
\(=\left(2x+y\right)^2\)
f) x^2 + 4xy + 4y^2
\(=x^2+4xy+\left(2y\right)^2\)
\(=\left(x+2y\right)^2\)
p/s: --.--
Bài 1:
\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)
Bài 2:
\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)
a) \(x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)
b) \(5x\left(3x-2\right)=4-9x^2\)
\(5x\left(3x-2\right)-\left(4-9x^2\right)=0\)
\(5x\left(3x-2\right)-\left(2-3x\right)\left(2+3x\right)=0\)
\(5x\left(3x-2\right)+\left(3x-2\right)\left(2+3x\right)=0\)
\(\left(3x-2\right)\left(5x+3x+2\right)=0\)
\(\left(3x-2\right)\left(8x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}}\)
c) \(x^2+7x=8\)
\(x^2+7x-8=0\)
\(x^2+8x-x-8=0\)
\(x\left(x+8\right)-\left(x+8\right)=0\)
\(\left(x+8\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+8=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-8\\x=1\end{cases}}}\)
d) \(2x^2+4y^2+10x+4xy=-25\)
\(x^2+x^2+4y^2+10x+4xy+25=0\)
\(\left(4y^2+4xy+x^2\right)+\left(x^2+10x+25\right)=0\)
\(\left(2y+x\right)^2+\left(x+5\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2y+x=0\\x+5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{5}{2}\\x=-5\end{cases}}}\)
a)\(8x^2+4xy=4x\left(2x+y\right)\)
Xin lỗi mình chỉ biết làm câu a
Chào bạn, sau đây là câu tl của mk:
a) 8x2 + 4xy = 4x.(2x + y)
b) x3 + 27 - 3x.(x + 3) = (x + 3)(x2 - 3x + 9) - 3x.(x + 3)
= (x + 3)(x2 - 3x + 9 - 3x)
= (x + 3)(x2 - 6x + 9)
= (x + 3)(x2 - 3x - 3x + 9)
= (x + 3)[x(x - 3) - 3(x - 3)]
= (x + 3)(x - 3)2
c) x3 + 2x2 + xy2 - 9x = x(x2 + 2x + y2 - 9)
d) 2x2 - 50 +2y2 + 4xy = (2x2+ 4xy + 2y2) - 50
= 2(x2 + 2xy + y2) - 2.25
= 2.(x + y)2 - 2.52 = 2.[(x + y)2 - 52]
= x.(x + y + 5)(x + y - 5)
Chúc bạn học tốt! 
a, \(x^2-4x-5\)
\(=x^2-5x+x-5=x\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+1\right)\left(x-5\right)\)
b, \(x^2-4xy+4y^2-36\)
\(=\left(x-2y\right)^2-36=\left(x-2y-6\right)\left(x-2y+6\right)\)
c, \(x^2-9x+20=x^2-5x-4x+20\)
\(=x\left(x-5\right)-4\left(x-5\right)\)
\(=\left(x-4\right)\left(x-5\right)\)
d, \(\left(9x^2-36\right)-\left(3x-6\right)\left(2x+5\right)\)
\(=\left(3x-6\right)\left(3x+6\right)-\left(3x-6\right)\left(2x+5\right)\)
\(=\left(3x-6\right)\left(3x+6-2x-5\right)\)
\(=3\left(x-2\right)\left(x+1\right)\)
a) \(x^2-4x-5=x^2-4x+4-9\)
= \(\left(x-2\right)^2-3^2\)
= \(\left(x-2-3\right)\left(x-2+3\right)=\left(x-5\right)\left(x+1\right)\)
b) \(x^2-4xy+4y^2-36\)
= \(\left(x-2y\right)^2-6^2=\left(x-2y-6\right)\left(x-2y+6\right)\)
c) \(x^2-9x+20\)
= \(\left(x-5\right)\left(x-4\right)\)
d) \(\left(9x^2-36\right)-\left(3x-6\right)\left(2x+5\right)\)
= \(\left(3x-6\right)\left(3x+6\right)-\left(3x-6\right)\left(2x+5\right)\)
= \(\left(3x-6\right)\left(3x+6-2x-5\right)=\left(3x-6\right)\left(x+1\right)\)