mọi người giải chi tiết giúp mik với ạ

Đáp án+Giải thích các bước giải:

Bài 1:

$a) x^3+2x=x(x^2+2)$

$b)3x-6y=3(x-2y)$

$c)5(x+3y)-15x(x+3y)=(x+3y)(5-15x)=5(x+3y)(1-3x)$

$d) 3(x-y)-5x(y-x)=3(x-y)+5x(x-y)=(x-y)(3+5x)$

$e) 2x+2y=2(x+y)$

$f) 2x+4y-6z=2(x+2y-3z)$

$g) -2x^2y-4xy^2-6xy=-2xy(x+2y+3)$

$h) 3a^2y-6a^2y+12a = 3a(ay-2ay+4)$

Bài 2:

$a) 4x^2-6x = 2x(2x-3)$

$b) x^3y-2x^2y^2+5xy=xy(x^2-2xy+5)$

$c) 2x^2(x+1)+4x(x+1)=(x+1)(2x^2+4x)=2x(x+1)(x+2)$

$d) \dfrac{2}{5}x(y-1)-\dfrac{2}{5}y(1-y)= \dfrac{2}{5}x(y-1)+\dfrac{2}{5}y(y-1)=\dfrac{2}{5}(y-1)(x+y)$

$e) 2(x-1)^3-5(x-1)^2-(x-1)=(x-1)[2(x-1)^2-5(x-1)-1]$

$f) x(y-x)^3-y(x-y)^2+xy(x-y)=-x(x-y)^3-y(x-y)^2+xy(x-y)=(x-y)[-x(x-y)^2-y(x-y)+xy]$

$g) xy(x+y)-2x-2y=xy(x+y)-2(x+y)=(x+y)(xy-2)$

$h) x(x+y)^2-y(x+y)^2-y^2(x-y)= (x+y)^2(x-y)-y^2(x-y)=(x-y)[(x+y)^2-y^2]=x(x-y)(x+2y)$

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Đáp án:

Kiến thức được áp dụng:

`(+)` Áp dụng phương pháp tách nhân tử chung để giải.

`(+)` Áp dụng phương pháp nhóm nhân tử chung.

`(+)` Áp dụng phương pháp tách nhân tử chung.

Bài toán 1:

a) `x^2+2x=x(x+2)`

b) `3x-6y=3(x-3y)`

c) `5(x+3y)-15x(x+3y)`

`=(5-15x)(x+3y)`

`=5(1-3x)(x+3y)`

d) `3(x-y)-5x(y-x)`

`=3(x-y)+5x(x-y)`

`=(3+5x)(x-y)`

e) `2x+2y`

`=2(x+y)`

f) `2x+4y-6z`

`=2(x+2y-3z)`

g) `-2x^2y-4xy^2-6xy`

`=-2xy(x+2y+3)`

h) `3a^2y-6a^2y+12a`

`=3a(ay-2ay+4)`

Bài toán 2: 

a) `4x^2-6x`

`=2x(2x-3)`

b) `x^3y-2x^2y^2+5xy`

`=xy(x^2-2xy+5)`

c) `2x^2(x+1)+4x(x+1)`

`=(2x^2+4x)(x+1)`

`=2x(x+2)(x+1)`

d) `2/5x(y-1)-2/5y(1-y)`

`=2/5x(y-1)+2/5y(y-1)`

`=(y-1)(2/5x+2/5y)`

`=2/5(y-1)(x+y)`

e) `2(x-1)^3-5(x-1)^2-(x-1)`

`=(x-1)[2(x-1)^2 -5(x-1)-1]`

`=(x-1)[(2(x^2-2x+1)-5x+5-1]`

`=(x-1)(2x^2-4x+2-5x+4)`

`=(x-1)(2x^2-9x+6)`

f) `x(y-x)^3-y(x-y)^2+xy(x-y)`

`=x(y-x)^3-y(y-x)^2-xy(y-x)`

`=(y-x)[x(y-x)^2-y(y-x)-xy]`

`=(y-x)[x(y^2-2xy+x^2)-y^2+xy-xy]`

`=(y-x)(xy^2-2x^2y+x^3-y^2)`

`=(y-x)(x^3-2x^2y+xy^2-y^2)`

g) `xy(x+y)-2x-2y`

`=xy(x+y)-2(x+y)`

`=(xy-2)(x+y)`

f) `x(x+y)^2-y(x+y)^2-y^2(x-y)`

`=(x+y)^2(x-y)-y^2(x-y)`

`=[(x+y)^2-y^2](x-y)`

`=x(x+2y)(x-y)`