`2)`
`a)VP=(a-b)³+3ab(a-b)`
`=a³-3a²b+3ab²-b³+3a²b-3ab²`
`=a³+(-3a²b+3a²b)+(3ab²-3ab²)-b³`
`=a³-b³`
`=VT`
`⇒đpcm`
`b)` Sửa đề:`(x+y²)-y²=x(x+2y)`
`→(x+y)²-y²=x(x+2y)`
`VT=(x+y)²-y²`
`=x²+2xy+y²-y²`
`=(x²+2xy)+(y²-y²)`
`=x(x+2y)`
`=VP`
`⇒đpcm`
`c)VP=x(x-3y)²+y(y-3x)²`
`=x(x²-6xy+9y²)+y(y²-6xy+9x²)`
`=x³-6x²y+9xy²+y³-6xy²+9x²y`
`=x³+(-6x²y+9x²y)+(9xy²-6xy²)+y³`
`=x³+3x²y+3xy²+y³`
`=(x+y)³`
`=VT`
`⇒đpcm`
`3)`
`a)x²-6x+10`
`=x²-6x+9+1`
`=(x²-6x+9)+1`
`=(x²-2.x.3+3²)+1`
`=(x-3)²+1`
Ta có:`(x-3)²≥0∀x`
`⇒(x-3)²+1≥1>0∀x`
`⇒x²-6x+10>0∀x(đpcm)`
`b)x²-2x+5`
`=x²-2x+1+4`
`=(x²-2x+1)+4`
`=(x²-2.x.1+1²)+4`
`=(x-1)²+4`
Ta có:`(x-1)²≥0∀x`
`⇒(x-1)²+4≥4>0∀x`
`⇒x²-2x+5>0∀x(đpcm)`
`c)4x-x²-5`
`=-(x²-4x+5)`
`=-(x²-4x+4+1)`
`=-(x²-4x+4)-1`
`=-(x²-2.x.2+2²)-1`
`=-(x-2)²-1`
Ta có:`(x-2)²≥0∀x`
`⇒-(x-2)²≤0∀x`
`⇒-(x-2)²-1≤-1<0∀x`
`⇒4x-x²-5<0∀x(đpcm)`
