Câu hỏi của Minh Thư Nguyễn

Question

tìm x$\in$ Qa) $\left(2x-1\right)^3=-27$b) $\left(2x-1\right)^6=\left(2x-1\right)^8$

___Kiều My___ · Accepted Answer

a) $\left(2x-1\right)^3=-27$$\left(2x-1\right)^3=-3^3$$2x-1=-3$$2x=-3+1$$2x=-2$$x=-2:2$$x=-1$b) $\left(2x-1\right)^6=\left(2x-1\right)^8$*$\Rightarrow2x-1=1$$2x=1+1$$2x=2$$x=2:2$$x=1$*$\Rightarrow2x-1=-1$$2x=-1+1$$2x=0$$x=0:2$$x=0$*$\Rightarrow2x-1=0$$2x=0+1$$2x=1$$x=1:2$$x=\frac{1}{2}$Vậy $x=\left\{1;0;\frac{1}{2}\right\}$Câu trả lời: a) $\left(2x-1\right)^3=-27$$\left(2x-1\right)^3=-3^3$$2x-1=-3$$2x=-3+1$$2x=-2$$x=-2:2$$x=-1$b) $\left(2x-1\right)^6=\left(2x-1\right)^8$*$\Rightarrow2x-1=1$$2x=1+1$$2x=2$$x=2:2$$x=1$*$\Rightarrow2x-1=-1$$2x=-1+1$$2x=0$$x=0:2$$x=0$*$\Rightarrow2x-1=0$$2x=0+1$$2x=1$$x=1:2$$x=\frac{1}{2}$Vậy $x=\left\{1;0;\frac{1}{2}\right\}$

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