Đáp án:
`a)x∈{`$\sqrt[]{10}+3$`;`$3-\sqrt[]{10}$`}`
`b)x=25/9`
Giải thích các bước giải:
`a)(2x-1)²-(3+x)²=(x-2)(x+2)+2(x-1)`
`⇔4x²-4x+1-(9+6x+x²)=x²-4+2x-2`
`⇔4x²-4x+1-9-6x-x²=x²-4+2x-2`
`⇔4x²-4x+1-9-6x-x²-x²+4-2x+2=0`
`⇔(4x²-x²-x²)-(4x+6x+2x)+(1-9+4+2)=0`
`⇔2x²-12x-2=0`
`⇔2(x²-6x-1)=0`
`⇔x²-6x-1=0`
`⇔x²-6x+9-10=0`
`⇔x²-6x+9=10`
`⇔(x-3)²=(`$\sqrt[]{10}$ `)^2`
`⇔`$\left[\begin{matrix} x-3=\sqrt[]{10}\\ x-3=-\sqrt[]{10}\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=\sqrt[]{10}+3\\ x=3-\sqrt[]{10}\end{matrix}\right.$
Vậy `x∈{`$\sqrt[]{10}+3$`;`$3-\sqrt[]{10}$`}`
`b)(3-2x)²-(x-2)²=2(x-3)(x+3)+(x-1)(x+2)`
`⇔9-12x+4x²-(x²-4x+4)=2(x²-9)+x²+2x-x-2`
`⇔9-12x+4x²-x²+4x-4=2x²-18+x²+2x-x-2`
`⇔(4x²-x²)+(-12x+4x)+(9-4)=(2x²+x²)+(2x-x)-(18+2)`
`⇔3x²-8x+5=3x²+x-20`
`⇔3x²-8x-3x²-x=-20-5`
`⇔-9x=-25`
`⇔x=25/9`
Vậy `x=25/9`
