Kiến thức toán lớp 8 bài 7

`1)`

`a)` `5.x-2.x.y=x.(5-2y)`

`b)` `5x.(x-1)-3x.(x-1)=(5x-3x).(x-1)=2x.(x-1)`

`c)` `x.(x+y)-5x-5y`

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

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

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

`d)` `x^2-9=x^2-3^2=(x-3).(x+3)`

`e)` `4x^2-25=(2x)^2-5^2=(2x-5).(2x+5)`

`f)` `x^6-y^6`

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

`=(x^3-y^3).(x^3+y^3)`

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

`g)` `9x^2+6xy+y^2=(3x)^2+2.3x.y+y^2=(3x+y)^2`

`h)` `6x-9-x^2`

`=-x^2+6x-9`

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

`=-(x^2-2.x.3+3^2)`

`=-(x-3)^2`

`l)` `x^2+4xy+4y^2=x^2+2.x.2y+(2y)^2=(x+2y)^2`

`i)` `(3x+1)^2-(x+1)^2`

`=(3x+1-x-1).(3x+1+x+1)`

`=2x.(4x+2)`

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

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

`2)`

`a)` `x+5x^2=0`

`<=>x.(1+5x)=0`

Trường hợp `1:`  `x=0`

Trường hợp `2:` 

`1+5x=0`

`<=>5x=0-1`

`<=>5x=-1`

`<=>x=-1/5`

Vậy `x in {0;-1/5}`

`b)` `x^3+x=0`

`<=>x.(x^2+1)=0`

Trường hợp `1:` `x=0`

Trường hợp `2:`

`x^2+1=0`

`<=>x^2=-1` ( vô lý vì `x^2ge0AAx` )

Vậy `x=0`

`c)` `x^3-0,25x=0`

`<=>x^3-1/4x=0`

`<=>x.(x^2-1/4)=0`

Trường hợp `1:` `x=0`

Trường hợp `2:`

`x^2-1/4=0`

`<=>x^2=1/4`

`<=>` `x=+-1/2`

Vậy `x in {0;1/2;-1/2}`

`d)` `x^2-10x=-25`

`<=>x^2-10x+25=0`

`<=>x^2-2.x.5+5^2=0`

`<=>(x-5)^2=0`

`<=>x-5=0`

`<=>x=5`

Vậy `x=5`

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

`<=>x+1-(x+1)^2=0`

`<=>1.(x+1)-(x+1).(x+1)=0`

`<=>(x+1).[1-(x+1)]=0`

`<=>(x+1).(-x)=0`

Trường hợp `1:`

`-x=0`

`<=>x=0`

Trường hợp `2:`

`x+1=0`

`=>x=0-1`

`=>x=-1`

Vậy `x in {-1;0}`

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