Bài 1:
`a)x²+2x+1=0`
`⇔x²+2.x.1+1²=0`
`⇔(x+1)²=0`
`⇔x+1=0`
`⇔x=-1`
Vậy `x=-1`
`b)16x²-8x+1=0`
`⇔(4x)²-2.4x.1+1²=0`
`⇔(4x-1)²=0`
`⇔4x-1=0`
`⇔4x=1`
`⇔x=1/4`
Vậy `x=1/4`
`c)1-4x+4x²=0`
`⇔1²-2.1.2x+(2x)²=0`
`⇔(1-2x)²=0`
`⇔1-2x=0`
`⇔2x=1`
`⇔x=1/2`
Vậy `x=1/2`
`d)25x²+20x+4=0`
`⇔(5x)²+2.5x.2+2²=0`
`⇔(5x+2)²=0`
`⇔5x+2=0`
`⇔5x=-2`
`⇔x=-2/5`
Vậy `x=-2/5`
Bài 2:
`a)A=(3x+1)(x-2)-3(x-1)(x+1)`
`=3x²-6x+x-2-3(x²-1)`
`=3x²-6x+x-2-3x²+3`
`=(3x²-3x²)+(-6x+x)+(-2+3)`
`=-5x+1`
`b)B=(2x-5)²+(10-2x)²+2(2x-5)(10-2x)`
`=(2x-5)²+2(2x-5)(10-2x)+(10-2x)²`
`=(2x-5+10-2x)²`
`=5²`
`=25`
`c)C=(x+7)(x²-7x+49)-(x-7)(x²+7x-49)`
`=(x+7)(x²-x.7+7²)-(x³+7x²-49x-7x²-49x+343)`
`=x³+7³-x³-7x²+49x+7x²+49x-343`
`=x³+343-x³-7x²+49x+7x²+49x-343`
`=(x³-x³)+(-7x²+7x²)+(49x+49x)+(343-343)`
`=98x`
`----------------`
Sửa đề:`C=(x+7)(x²-7x+49)-(x-7)(x²+7x-49)`
`→C=(x+7)(x²-7x+49)-(x-7)(x²+7x+49)`
`C=(x+7)(x²-7x+49)-(x-7)(x²+7x+49)`
`=(x+7)(x²-x.7+7²)-(x-7)(x²+x.7+7²)`
`=x³+7³-(x³-7³)`
`=x³+343-(x³-343)`
`=x³+343-x³+343`
`=(x³-x³)+(343+343)`
`=686`
`----------------`
`d)D=(5x+1)(5x-1)-25x²+11`
`=(5x)²-1²-25x²+11`
`=25x²-1-25x²+11`
`=(25x²-25x²)+(-1+11)`
`=10`
