`1)`
`a)(2x+1)²-4(x+2)²=9`
`⇔4x²+4x+1-4(x²+4x+4)=9`
`⇔4x²+4x+1-4x²-16x-16=9`
`⇔(4x²-4x²)+(4x-16x)+(1-16)=9`
`⇔-12x-15=9`
`⇔-12x=9+15`
`⇔-12x=14`
`⇔x=24:(-12)`
`⇔x=-2`
Vậy `x=-2`
`b)(3x-1)²+2(x+3)²+11(x+1)(1-x)=6`
`⇔9x²-6x+1+2(x²+6x+9)+11(1-x²)=6`
`⇔9x²-6x+1+2x²+12x+18+11-11x²=6`
`⇔(9x²+2x²-11x²)+(-6x+12x)+(1+18+11)=6`
`⇔6x+30=6`
`⇔6x=6-30`
`⇔6x=-24`
`⇔x=(-24):6`
`⇔x=-4`
Vậy `x=-4`
`c)(x+1)³-x²(x+3)=2`
`⇔x³+3x²+3x+1-x³-3x²=2`
`⇔(x³-x³)+(3x²-3x²)+3x+1=2`
`⇔3x+1=2`
`⇔3x=2-1`
`⇔3x=1`
`⇔x=1/3`
Vậy `x=1/3`
`d)(x-2)³-x(x+1)(x-1)+6x²=5`
`⇔x³-6x²+12x-8-x(x²-1)+6x²=5`
`⇔x³-6x²+12x-8-x³+x+6x²=5`
`⇔(x³-x³)+(-6x²+6x²)+(12x+x)-8=5`
`⇔13x-8=5`
`⇔13x=5+8`
`⇔13x=13`
`⇔x=13:13`
`⇔x=1`
Vậy `x=1`
`e)(x-3)(x²+3x+9)-x(x+4)(x-4)=5`
`⇔x³-27-x(x²-16)=5`
`⇔x³-27-x³+16x=5`
`⇔(x³-x³)+16x-27=5`
`⇔16x-27=5`
`⇔16x=5+27`
`⇔16x=32`
`⇔x=32:16`
`⇔x=2`
Vậy `x=2`
`f)(x-2)³-(x+5)(x²-5x+25)+6x²=11`
`⇔x³-6x²+12x-8-(x³+125)+6x²=11`
`⇔x³-6x²+12x-8-x³-125+6x²=11`
`⇔(x³-x³)+(-6x²+6x²)+12x-(8+125)=11`
`⇔12x-133=11`
`⇔12x=11+133`
`⇔12x=144`
`⇔x=144:12`
`⇔x=12`
Vậy `x=12`
`2)`
`a)A=(2x+y)(4x²-2xy+y²)-(2x-y)(4x²+2xy+y²)`
`=(2x+y)[(2x)²-2x.y+y²]-(2x-y)[(2x)²+2x.y+y²]`
`=(2x)³+y³-[(2x)³-y³]`
`=8x³+y³-(8x³-y³)`
`=8x³+y³-8x³+y³`
`=(8x³-8x³)+(y³+y³)`
`=2y³`
`b)B=(x+3)(x²-3x+9)`
`=(x+3)(x²-x.3+3²)`
`=x³+3³`
`=x³+27`
`c)C=2(x+y)(x-y)+(x+y)²+(x-y)²`
`=(x+y)²+2(x+y)(x-y)+(x-y)²`
`=(x+y+x-y)²`
`=(2x)²`
`=4x²`
`d)D=(x-y+z)²+(z-y)²+2(x-y+z)(y-z)`
`=(x-y+z)²+2(x-y+z)(y-z)+(y-z)²`
`=(x-y+z+y-z)²`
`=x²`
`e)E=(5x²-1)^2+2(1-5x)(4+5x)+(5x+4)²`
`=25x^4-10x²+1+(2-10x)(4+5x)+25x²+40x+16`
`=25x^4-10x²+1+8+10x-40x-50x²+25x²+40x+16`
`=25x^4+(-10x²-50x²+25x²)+(10x-40x+40x)+(1+8+16)`
`=25x^4-35x²+10x+25`
Sửa đề:`E=(5x²-1)^2+2(1-5x)(4+5x)+(5x+4)²`
`→E=(5x-1)^2+2(1-5x)(4+5x)+(5x+4)²`
`E=(5x-1)^2+2(1-5x)(4+5x)+(5x+4)²`
`=(1-5x)²+2(1-5x)(5x+4)+(5x+4)²`
`=(1-5x+5x+4)²`
`=5²`
`=25`
