help meeeeeeeeeeeeeeeeeeeeeee
Hãy luôn nhớ cảm ơn và vote 5*
nếu câu trả lời hữu ích nhé!
`1)`
`x^2-2x+1-y^2+2x-1`
`=(x^2-2x+1)-(y^2-2x+1)`
`=(x-1)^2-(y-1)^2`
`=(x-1-y+1)(x-1+y-1)`
`=(x-y)(x+y-2)`
`2)`
`x^2-4x+4-y^2-6y-9`
`=(x^2-4x+4)-(y^2+6y+9)`
`=(x-2)^2-(y+3)^2`
`=(x-2-y-3)(x-2+y+3)`
`=(x-y-5)(x+y+1)`
`3)`
`4x^2-4x+1-y^2-8y-16`
`=(4x^2-4x+1)-(y^2+8y+16)`
`=(2x-1)^2-(y+4)^2`
`=(2x-1-y-4)(2x-1+y+4)`
`=(2x-y-5)(2x+y+3)`
`4)`
`x^2-2xy+y^2-z^2+2zt-t^2`
`=(x^2-2xy+y^2)-(z^2-2zt+t^2)`
`=(x-y)^2-(z-t)^2`
`=(x-y-z+t)(x-y+z-t)`
`5)`
`(x+y)^2-2(x+y)+1`
`=(x+y)^2-2·(x+y)·1+1^2`
`=(x+y+1)^2`
`6)`
`(x^2+x+1)^2+2x(x^2+x+1)+x^2`
`=(x^2+x+1)^2+2·(x^2+x+1)·x+x^2`
`=(x^2+x+1+x)^2`
`=(x^2+2x+1)^2`
`=[(x+1)^2]^2`
`=(x+1)^4`
Hãy giúp mọi người biết câu trả lời này thế nào?
Đáp án `:`
`1)`
`x^2-2x+1-y^2+2x-1`
`=(x^2-2x+1)-(y^2-2x+1)`
`=(x-1)^2-(y-1)^2`
`=[(x-1)-(y-1)][(x-1)+(y-1)]`
`=(x-1-y+1)(x-1+y-1)`
`=(x-y)(x+y-2)`
`2)`
`x^2-4x+4-y^2-6y-9`
`=(x^2-4x+4)-(y^2+6y+9)`
`=(x-2)^2-(y+3)^2`
`=(x-2-y-3)(x-2+y+3)`
`=(x-y-5)(x+y+1)`
`3)`
`4 x^2-4x+1-y^2-8y-16`
`=(4 x^2-4x+1)-(y^2+8y+16)`
`=(2x-1)^2-(y+4)^2`
`=(2x-1-y-4)(2x-1+y+4)`
`=(2x-y-5)(2x+y+3)`
`4)`
`x^2-2xy+y^2-z^2+2zt-t^2`
`=(x^2-2xy+y^2)-(z^2-2zt+t^2)`
`=(x-y)^2-(z-t)^2`
`=[(x-y)-(z-t)][(x-y)+(z-t)]`
`=(x-y-z+t)(x-y+z-t)`
`5)`
`(x+y)^2-2(x+y)+1`
`=(x+y)^2-2(x+y).1+1^2`
`=(x+y-1)^2`
`6)`
`(x^2+x+1)^2+2x(x^2+x+1)+x^2`
`=(x^2+x+1+x)^2`
`=(x^2+2x+1)^2`
`=[(x+1)^2]^2=(x+1)^4`
Hãy giúp mọi người biết câu trả lời này thế nào?
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