Đáp án:
`a)x=8/9`
`b)x∈{`$\dfrac{\sqrt[]{265}+5}{12}$`;`$\dfrac{5-\sqrt[]{265}}{12}$`}`
`c)x=-1`
Giải thích các bước giải:
`a)3x²(2x-7)+(2-x)(6x²-x+3)+8x²=4x-2`
`⇔6x³-21x²+12x²-2x+6-6x³+x²-3x+8x²=4x-2`
`⇔(6x³-6x³)+(-21x²+12x²+x²+8x²)+(-2x-3x)+6=4x-2`
`⇔-5x+6=4x-2`
`⇔-5x-4x=-2-6`
`⇔-9x=-8`
`⇔x=8/9`
Vậy `x=8/9`
`b)6x²-(2x+5)-(3x-2)=7`
`⇔6x²-2x-5-3x+2=7`
`⇔6x²-(2x+3x)-(5-2)=7`
`⇔6x²-5x-3=7`
`⇔6x²-5x-3-7=0`
`⇔6x²-5x-10=0`
`⇔6(x²-5/6x-5/3)=0`
`⇔6(x²-5/6x+25/144-265/144)=0`
`⇔6(x²-5/6x+25/144)-265/24=0`
`⇔6[x²-2.x. 5/12+(5/12)^2]=265/24`
`⇔6(x-5/12)^2=265/24`
`⇔(x-5/12)^2=265/24:6`
`⇔(x-5/12)^2=265/24 . 1/6`
`⇔(x-5/12)^2=265/144`
`⇔(x-5/12)^2=(`$\dfrac{\sqrt[]{265}}{12}$`)^2`
`⇔`$\left[\begin{matrix} x-\dfrac{5}{12}=\dfrac{\sqrt[]{265}}{12}\\ x-\dfrac{5}{12}=-\dfrac{\sqrt[]{265}}{12}\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=\dfrac{\sqrt[]{265}}{12}+\dfrac{5}{12}\\ x=-\dfrac{\sqrt[]{265}}{12}+\dfrac{5}{12}\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=\dfrac{\sqrt[]{265}+5}{12}\\ x=\dfrac{5-\sqrt[]{265}}{12}\end{matrix}\right.$
Vậy `x∈{`$\dfrac{\sqrt[]{265}+5}{12}$`;`$\dfrac{5-\sqrt[]{265}}{12}$`}`
`c)x(x-5)(x+5)-(x+2)(x²-2x+4)=17`
`⇔x(x²-25)-(x³+8)=17`
`⇔x³-25x-x³-8=17`
`⇔-25x-8=17`
`⇔-25x=17+8`
`⇔-25x=25`
`⇔x=25:(-25)`
`⇔x=-1`
Vậy `x=-1`
