Bài 1:
`1)3x(12x-4)-9x(4x-3)=30`
`⇔36x²-12x-36x²+27x=30`
`⇔(36x²-36x²)+(-12x+27x)=30`
`⇔15x=30`
`⇔x=30:15`
`⇔x=2`
Vậy `S={2}`
`2)(5-2x).x+2x(x-1)=15`
`⇔5x-2x²+2x²-2x=15`
`⇔(5x-2x)+(-2x²+2x²)=15`
`⇔3x=15`
`⇔x=15:3`
`⇔x=5`
Vậy `S={5}`
`3)(x-2)³-(x-1)(x+2)-x²+(x+2)³=2x³`
`⇔x³-6x²+12x-8-(x²+2x-x-2)-x²+x³+6x²+12x+8=2x³`
`⇔x³-6x²+12x-8-x²-2x+x+2-x²+x³+6x²+12x+8-2x³=0`
`⇔(x³+x³-2x³)+(-6x²-x²-x²+6x²)+(12x-2x+x+12x)+(-8+2+8)=0`
`⇔-2x²+23x+2=0`
`⇔-2(x²-23/2x-1)=0`
`⇔-2(x²-23/2x+529/16-545/16)=0`
`⇔-2(x²-23/2x+529/16)+545/8=0`
`⇔-2[x²-2.x. 23/4+(23/4)^2]=-545/8`
`⇔-2(x-23/4)^2=-545/8`
`⇔(x-23/4)^2=(-545/8):(-2)`
`⇔(x-23/4)^2=(-545/8). (-1/2)`
`⇔(x-23/4)^2=545/16`
`⇔(x-23/4)^2=(`$\dfrac{\sqrt[]{545}}{4}$`)^2`
`⇔`\(\left[ \begin{array}{l}x-\dfrac{23}{4}=\dfrac{\sqrt[]{545}}{4}\\x-\dfrac{23}{4}=-\dfrac{\sqrt[]{545}}{4}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt[]{545}}{4}+\dfrac{23}{4}\\x=-\dfrac{\sqrt[]{545}}{4}+\dfrac{23}{4}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt[]{545}+23}{4}\\x=\dfrac{23-\sqrt[]{545}}{4}\end{array} \right.\)
Vậy `S={`$\dfrac{\sqrt[]{545}+23}{4}$`;`$\dfrac{23-\sqrt[]{545}}{4}$`}`
`4)x³-3x²+3x-1=0`
`⇔x³-3.x².1+3.x.1²-1³=0`
`⇔(x-1)³=0`
`⇔x-1=0`
`⇔x=1`
Vậy `S={1}`
