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c,x(x2-2x+1)
x(x-1)2
d,(x3-1)-(3x2-3x)
=(x-1).(x2-x+1)-3x(x-1)
(x-1).(x2-x+1-3x)
=(x-1).(x2-4x+1)
a, 4x2 - 12x + 9
= (2x + 3)2
b, 9x4y3 + 3x2y4
= 3x2y3(3x2 + y)
c, ( x - 3 )2 - 2x ( x - 3 )
= (x - 3)(x - 3 - 2x)
= (x - 3)(-x - 3)
d, 3x ( x - 1 ) + 6 ( x - 1 )
= 3(x - 1)(x + 2)
e, 2x ( x + 1 ) - 4x - 4
= 2x(x + 1) - 4(x + 1)
= (x + 1)(2x - 4)
= 2(x + 1)(x - 2)
f, ( 2x - 3 )2 - 4x + 6
= (2x - 3)2 - 2(2x - 3)
= (2x - 3)(2x - 3 - 2)
= (2x - 3)(2x - 5)
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) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
a/ \(=3y^2-6y-2x+1\)
b/ \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
c/ \(=\left(2-x\right)^3\)
d/ \(=xy^2+x^2y+3xy+x^2y+x^3+3x^2-3xy-3x^2-9x\)
\(=xy\left(y+x+3\right)+x^2\left(y+x+3\right)-3x\left(y+x+3\right)\)
\(=\left(xy+x^2-3x\right)\left(y+x+3\right)=x\left(y+x-3\right)\left(y+x+3\right)\)
e/ \(=xy-x^2+2x-y^2+xy-2y\)
\(=x\left(y-x+2\right)-y\left(y-x+2\right)=\left(x-y\right)\left(y-x+2\right)\)
a) =(2x+3y-1)2
b)=-(x-1)3
c)=-(x3-6x2+12x-8)=-(x-2)3
d)x3 + 2x2y + xy2 – 9x
= x(x2 + 2xy + y2 -9)
= x[(x2 + 2xy + y2) - 32]
= x[(x + y)2 - 32]
= x (x + y – 3)(x + y + 3)
e) 2x-2y-x2+2xy-y2=2(x-y)-(x-y)2=(x-y)(2-x+y)
\(a,36-12x+x^2\)\(=x^2-2.x.6+6^2=\left(x-6\right)^2\)
\(b,4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2=\left(2x+3\right)^2\)
\(c,x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(d,2x^2-2y^2-6x-6y\)
\(=2\left(x-y\right)\left(x+y\right)-6\left(x+y\right)\)
\(=\left(x+y\right)\left(2x-2y-6\right)\)
\(e,x^3+3x^2-3x-1\)
\(=\left(x^3-1\right)+\left(3x^2-3x\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+3x\right)=\left(x-1\right)\left(x^2+4x+1\right)\)
cảm ơn bạn nha