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a) \(x^2-6x+9-9y^2=x^2-2\cdot3+3^2-\left(3y\right)^2=\left(x-3\right)^2-\left(3y\right)^2=\left(x-3-3y\right)\cdot\left(x-3+3y\right)\)
b) \(x^3-3x^2+2x-1+2\cdot\left(x^2-x\right)=\left(x-1\right)^3+2x\cdot\left(x-1\right)=\left(x-1\right)\cdot\left[\left(x-1\right)^2+2x\right]\)
a) (x2-4x+3)(x2-10x+24)+8=((x2-x)-(3x-3))((x2-6x)-(4x-24))+8
=(x(x-1)-3(x-1))(x(x-6)-4(x-6))+8=(x-1)(x-3)(x-4)(x-6)+8=((x-1)(x-6))(x-3)(x-4))+8
=(x2-7x+6)(x2-7x+12)+8
Đặt x2-7x+6=a
Ta có : a(a+6)+8=a2+6a+8=(a+2)(a+4)=(x2-7x+8)(x2-7x+10)=(x2-7x+8)(x-5)(x-2)
b) Tương tự như câu a kết quả là (x-3)(x3+9x2+21x+9)
c) x4+x3+6x2+3x+9=(x4+x3+3x2)+(3x2+3x+9)=x2(x2+x+3)+3(x2+x+3)=(x2+x+3)(x2+2)
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a: \(=3x^2+3x-x-1\)
=(x+1)(3x-1)
b: \(=x^3+x^2+5x^2+5x+6x+6\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x+2\right)\cdot\left(x+3\right)\)
c: \(=x^4+3x^2-x^2-3\)
\(=\left(x^2+3\right)\left(x^2-1\right)\)
\(=\left(x^2+3\right)\left(x-1\right)\left(x+1\right)\)
f: \(=5x\left(x^2+3x+2\right)\)
=5x(x+1)(x+2)
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
1)
$x^3+9x^2+23x+15=(x^3+x^2)+(8x^2+8x)+(15x+15)$
$=x^2(x+1)+8x(x+1)+15(x+1)$
$=(x+1)(x^2+8x+15)$
$=(x+1)[(x^2+3x)+(5x+15)]$
$=(x+1)[x(x+3)+5(x+3)]=(x+1)(x+3)(x+5)$
5)
$x^4+5x^2+9=(x^4+6x^2+9)-x^2$
$=(x^2+3)^2-x^2=(x^2+3-x)(x^2+3+x)$
3)
$(3x-2)^2(6x-5)(6x-3)-5$
$=(9x^2-12x+4)(36x^2-48x+15)-5$
$=(9x^2-12x+4)[4(9x^2-12x)+15]-5$
$=(a+4)(4a+15)-5$ (đặt $9x^2-12x=a$)
$=4a^2+31a+55$
$=4a^2+20a+11a+55$
$=4a(a+5)+11(a+5)=(4a+11)(a+5)=(36x^2-48x+11)(9x^2-12x+5)$
$=