Câu hỏi của Văn Nam Vương Nguyễn

Question

(1/3 - 3/2*x )mũ 2 = 9/4giải giúp em ạ em cảm ơn

Võ Ngọc Phương · Accepted Answer

$\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\dfrac{9}{4}$$\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{3}{2}\right)^2$ hoặc $\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{-3}{2}\right)^2$$=>\dfrac{1}{3}-\dfrac{3}{2}x=\dfrac{3}{2}$ hoặc $\dfrac{1}{3}-\dfrac{3}{2}x=\dfrac{-3}{2}$$\dfrac{3}{2}x=\dfrac{3}{2}+\dfrac{1}{3}$ hoặc $\dfrac{3}{2}x=\dfrac{-3}{2}+\dfrac{1}{3}$$\dfrac{3}{2}x=\dfrac{9}{6}+\dfrac{2}{6}$ hoặc $\dfrac{3}{2}x=-\dfrac{9}{6}+\dfrac{2}{6}$$\dfrac{3}{2}x=\dfrac{11}{6}$ hoặc $\dfrac{3}{2}x=\dfrac{-7}{6}$$x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}$ hoặc $x=\dfrac{-7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}$$x=\dfrac{11}{9}$ hoặc $x=-\dfrac{7}{9}$Vậy...Câu trả lời: $\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\dfrac{9}{4}$$\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{3}{2}\right)^2$ hoặc $\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{-3}{2}\right)^2$$=>\dfrac{1}{3}-\dfrac{3}{2}x=\dfrac{3}{2}$ hoặc $\dfrac{1}{3}-\dfrac{3}{2}x=\dfrac{-3}{2}$$\dfrac{3}{2}x=\dfrac{3}{2}+\dfrac{1}{3}$ hoặc $\dfrac{3}{2}x=\dfrac{-3}{2}+\dfrac{1}{3}$$\dfrac{3}{2}x=\dfrac{9}{6}+\dfrac{2}{6}$ hoặc $\dfrac{3}{2}x=-\dfrac{9}{6}+\dfrac{2}{6}$$\dfrac{3}{2}x=\dfrac{11}{6}$ hoặc $\dfrac{3}{2}x=\dfrac{-7}{6}$$x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}$ hoặc $x=\dfrac{-7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}$$x=\dfrac{11}{9}$ hoặc $x=-\dfrac{7}{9}$Vậy...

Võ Ngọc Phương · Answer

Mình sửa lại ( từ dòng 4 trở xuống ):... $\dfrac{3}{2}x=\dfrac{1}{3}-\dfrac{3}{2}$ hoặc $\dfrac{3}{2}x=\dfrac{1}{3}+\dfrac{3}{2}$$\dfrac{3}{2}x=\dfrac{2}{6}-\dfrac{9}{6}$ hoặc $\dfrac{3}{2}x=\dfrac{2}{6}+\dfrac{9}{6}$$\dfrac{3}{2}x=-\dfrac{7}{6}$ hoặc $\dfrac{3}{2}x=\dfrac{11}{6}$$x=-\dfrac{7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}$ hoặc $x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}$$x=\dfrac{-7}{9}$ hoặc $x=\dfrac{11}{9}$Vậy...Câu trả lời: Mình sửa lại ( từ dòng 4 trở xuống ):... $\dfrac{3}{2}x=\dfrac{1}{3}-\dfrac{3}{2}$ hoặc $\dfrac{3}{2}x=\dfrac{1}{3}+\dfrac{3}{2}$$\dfrac{3}{2}x=\dfrac{2}{6}-\dfrac{9}{6}$ hoặc $\dfrac{3}{2}x=\dfrac{2}{6}+\dfrac{9}{6}$$\dfrac{3}{2}x=-\dfrac{7}{6}$ hoặc $\dfrac{3}{2}x=\dfrac{11}{6}$$x=-\dfrac{7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}$ hoặc $x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}$$x=\dfrac{-7}{9}$ hoặc $x=\dfrac{11}{9}$Vậy...

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