Bài 4:
`1)(2x+1)²-2x(2x+3)=4`
`⇔4x²+4x+1-4x²-6x=4`
`⇔(4x²-4x²)+(4x-6x)+1=4`
`⇔-2x+1=4`
`⇔-2x=4-1`
`⇔-2x=3`
`⇔x=-3/2`
Vậy `x=-3/2`
`2)(x-3)(x+3)-(x-2)²=1`
`⇔x²-9-(x²-4x+4)=1`
`⇔x²-9-x²+4x-4=1`
`⇔(x²-x²)+4x-(9+4)=1`
`⇔4x-13=1`
`⇔4x=1+13`
`⇔4x=14`
`⇔x=14/4`
`⇔x=7/2`
Vậy `x=7/2`
`3)(x-1)(x²+x+1)-x(x²+2)-4=0`
`⇔x³-1-x³-2x-4=0`
`⇔(x³-x³)-2x-(1+4)=0`
`⇔-2x-5=0`
`⇔-2x=5`
`⇔x=-5/2`
Vậy `x=-5/2`
`4)(2-x)²-9=0`
`⇔(2-x)²-3²=0`
`⇔(2-x+3)(2-x-3)=0`
`⇔(5-x)(-x-1)=0`
`⇔`$\left[\begin{matrix} 5-x=0\\ -x-1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix}x=5\\x=-1\end{matrix}\right.$
Vậy `x∈{5;-1}`
`5)x(x-5)(x+5)-(x+2)(x²-2x+4)=3`
`⇔x(x²-25)-(x³+8)=3`
`⇔x³-25x-x³-8=3`
`⇔-25x-8=3`
`⇔-25x=3+8`
`⇔-25x=11`
`⇔x=-11/25`
Vậy `x=-11/25`
