Question

Answer 1

`1)`

`a)c^2-b^2+6b-9=c^2-(b-3)^2=(c-b+3)(c+b-3)`

`b)3x^3-12x^2+12x=3x(x^2-4x+4)=3x(x-2)^2`

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`2)`

`a)[x^2-2xy+y^2]/[x^2-xy]`   `ĐK:x \ne 0,x \ne y`

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

`b)[x^2-4]/[3x+6]`     `ĐK: x \ne -2`

`=[(x-2)(x+2)]/[3(x+2)]=[x-2]/3`

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`3)`

`a)x^2-x(x+2)-1=2`

`<=>x^2-x^2-2x=3`

`<=>-2x=3<=>x=-3/2`

Vậy `S={-3/2}`

 

`b)x^2=(2x-1)^2`

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

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

`<=>(1-x)(3x-1)=0`

`<=>[(x=1),(x=1/3):}`

Vậy `S={1;1/3}`