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2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
a) \(x^5+4x+5=\left(x^5+x^4\right)-\left(x^4+x^3\right)+\left(x^3+x^2\right)-\left(x^2+x\right)+\left(5x+5\right)=x^4\left(x+1\right)-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+5\left(x+1\right)=\left(x^4-x^3+x^2-x+5\right)\left(x+1\right)\)
b) \(x^4+6x^3+11x^2+6x+1=\left(x^4+3x^3+x^2\right)+\left(3x^3+9x^2+3x\right)+\left(x^2+3x+1\right)=x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)=\left(x^2+3x+1\right)^2\)
c) \(64x^4+1=\left[\left(8x^2\right)^2+16x^2+1\right]-16x^2=\left(8x^2+1\right)^2-\left(4x\right)^2=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)d) \(81x^4+4=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)
a: \(x^2+2x+1+4x+4\)
\(=\left(x^2+2x+1\right)+\left(4x+4\right)\)
\(=\left(x+1\right)^2+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x+1+4\right)\)
\(=\left(x+1\right)\left(x+5\right)\)
b: Sửa đề: \(2x^3+6x^2+x^2+3x\)
\(=2x^2\left(x+3\right)+x\left(x+3\right)\)
\(=\left(x+3\right)\left(2x^2+x\right)\)
\(=x\left(x+3\right)\left(2x+1\right)\)
c: \(\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{4}x+1\)
\(=\dfrac{1}{4}x\left(\dfrac{1}{4}x+1\right)+\left(\dfrac{1}{4}x+1\right)\)
\(=\left(\dfrac{1}{4}x+1\right)\left(\dfrac{1}{4}x+1\right)=\left(\dfrac{1}{4}x+1\right)^2\)
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
e) \(=x^2\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^2-2x+3\right)\)
g) \(=x^2\left(3x-1\right)-x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(x^2-x+4\right)\)
h) \(=3x^2\left(2x+1\right)-x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)
i) \(=2x^2\left(2x+1\right)+2x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(2x^2+2x+1\right)\)
