Câu hỏi của Đéo Còn Tên

Question

Bài 2:a)$x+\frac{1}{3}=\frac{3}{4}$                   b)$\left|x+\frac{1}{8}\right|-\frac{1}{6}=0$c)$\frac{x}{27}=-\frac{2}{36}$d)$\left(x+\frac{1}{2}\right)^2=\frac{1}{16}$

Nguyễn Huy Tú · Accepted Answer

Bài 2:a) $x+\frac{1}{3}=\frac{3}{4}$$\Rightarrow x=\frac{5}{12}$Vậy $x=\frac{5}{12}$b) $\left|x+\frac{1}{8}\right|-\frac{1}{6}=0$$\Rightarrow\left|x+\frac{1}{8}\right|=\frac{1}{6}$$\Rightarrow x+\frac{1}{8}=\frac{1}{6}$ hoặc $x+\frac{1}{8}=\frac{-1}{6}$+) $x+\frac{1}{8}=\frac{1}{6}\Rightarrow x=\frac{1}{24}$+) $x+\frac{1}{8}=\frac{-1}{6}\Rightarrow x=\frac{-7}{24}$Vậy $x\in\left\{\frac{1}{24};\frac{-7}{24}\right\}$c) $\frac{x}{27}=\frac{-2}{36}$$\Rightarrow\frac{x}{27}=\frac{-1}{18}$$\Rightarrow18x=-27$$\Rightarrow x=\frac{-3}{2}$Vậy $x=\frac{-3}{2}$d) $\left(x+\frac{1}{2}\right)^2=\frac{1}{16}$$\Rightarrow x+\frac{1}{2}=\frac{1}{4}$ hoặc $x+\frac{1}{2}=\frac{-1}{4}$+) $x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{-1}{4}$+) $x+\frac{1}{2}=\frac{-1}{4}\Rightarrow x=\frac{-3}{4}$Vậy $x\in\left\{\frac{-1}{4};\frac{-3}{4}\right\}$Câu trả lời: Bài 2:a) $x+\frac{1}{3}=\frac{3}{4}$$\Rightarrow x=\frac{5}{12}$Vậy $x=\frac{5}{12}$b) $\left|x+\frac{1}{8}\right|-\frac{1}{6}=0$$\Rightarrow\left|x+\frac{1}{8}\right|=\frac{1}{6}$$\Rightarrow x+\frac{1}{8}=\frac{1}{6}$ hoặc $x+\frac{1}{8}=\frac{-1}{6}$+) $x+\frac{1}{8}=\frac{1}{6}\Rightarrow x=\frac{1}{24}$+) $x+\frac{1}{8}=\frac{-1}{6}\Rightarrow x=\frac{-7}{24}$Vậy $x\in\left\{\frac{1}{24};\frac{-7}{24}\right\}$c) $\frac{x}{27}=\frac{-2}{36}$$\Rightarrow\frac{x}{27}=\frac{-1}{18}$$\Rightarrow18x=-27$$\Rightarrow x=\frac{-3}{2}$Vậy $x=\frac{-3}{2}$d) $\left(x+\frac{1}{2}\right)^2=\frac{1}{16}$$\Rightarrow x+\frac{1}{2}=\frac{1}{4}$ hoặc $x+\frac{1}{2}=\frac{-1}{4}$+) $x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{-1}{4}$+) $x+\frac{1}{2}=\frac{-1}{4}\Rightarrow x=\frac{-3}{4}$Vậy $x\in\left\{\frac{-1}{4};\frac{-3}{4}\right\}$

Lightning Farron · Answer

a)$x+\frac{1}{3}=\frac{3}{4}$$\Rightarrow x=\frac{3}{4}-\frac{1}{3}$$\Rightarrow x=\frac{5}{12}$b)$\left|x+\frac{1}{8}\right|-\frac{1}{6}=0$$\Rightarrow\left|x+\frac{1}{8}\right|=\frac{1}{6}$$\Rightarrow x+\frac{1}{8}=\frac{1}{6}$ hoặc $x+\frac{1}{8}=-\frac{1}{6}$$\Rightarrow x=\frac{1}{6}-\frac{1}{8}$ hoặc $x=-\frac{1}{6}-\frac{1}{8}$$\Rightarrow x=\frac{1}{24}$ hoặc $x=-\frac{7}{24}$c)$\frac{x}{27}=-\frac{2}{36}$$\Rightarrow x=\frac{\left(-2\right)\cdot27}{36}=-\frac{3}{2}$d)$\left(x+\frac{1}{2}\right)^2=\frac{1}{16}$$\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2$$\Rightarrow x+\frac{1}{2}=\frac{1}{4}$ hoặc $x+\frac{1}{2}=-\frac{1}{4}$$\Rightarrow x=\frac{1}{4}-\frac{1}{2}$ hoặc $x=-\frac{1}{4}-\frac{1}{2}$$\Rightarrow x=-\frac{1}{4}$ hoặc $x=-\frac{3}{4}$ Câu trả lời: a)$x+\frac{1}{3}=\frac{3}{4}$$\Rightarrow x=\frac{3}{4}-\frac{1}{3}$$\Rightarrow x=\frac{5}{12}$b)$\left|x+\frac{1}{8}\right|-\frac{1}{6}=0$$\Rightarrow\left|x+\frac{1}{8}\right|=\frac{1}{6}$$\Rightarrow x+\frac{1}{8}=\frac{1}{6}$ hoặc $x+\frac{1}{8}=-\frac{1}{6}$$\Rightarrow x=\frac{1}{6}-\frac{1}{8}$ hoặc $x=-\frac{1}{6}-\frac{1}{8}$$\Rightarrow x=\frac{1}{24}$ hoặc $x=-\frac{7}{24}$c)$\frac{x}{27}=-\frac{2}{36}$$\Rightarrow x=\frac{\left(-2\right)\cdot27}{36}=-\frac{3}{2}$d)$\left(x+\frac{1}{2}\right)^2=\frac{1}{16}$$\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2$$\Rightarrow x+\frac{1}{2}=\frac{1}{4}$ hoặc $x+\frac{1}{2}=-\frac{1}{4}$$\Rightarrow x=\frac{1}{4}-\frac{1}{2}$ hoặc $x=-\frac{1}{4}-\frac{1}{2}$$\Rightarrow x=-\frac{1}{4}$ hoặc $x=-\frac{3}{4}$

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