Đáp án:
\(6)\ x=3\\ 8)\ x=\dfrac{-7}{2}\\ 9)\ x=\dfrac{8}{3}\\ 10)\ x=\dfrac{1}{2}\ \text{hoặc}\ x=\dfrac{-1}{6}\\ 11)\ x=\dfrac{34}{15}\ \text{hoặc}\ x=\dfrac{-14}{15}\\ 12)\ x=\dfrac{1}{3}\ \text{hoặc}\ x=10\)
Giải thích các bước giải:
\(6)\ \bigg(-3\dfrac{1}{2}-2x\bigg):3\dfrac{2}{3}=\dfrac{15}{22}\\ ⇔\bigg(\dfrac{-7}{2}-2x\bigg):\dfrac{11}{3}=\dfrac{15}{22}\\ ⇔\dfrac{-7}{2}-2x=\dfrac{15}{22}.\dfrac{11}{3}\\ ⇔\dfrac{-7}{2}-2x=\dfrac{5}{2}\\ ⇔2x=\dfrac{-7}{2}-\dfrac{5}{2}\\ ⇔2x=-6\\ ⇔x=(-6):2=(-3)\\ \text{Vậy x = (- 3)}\\ 8)\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ ⇔\bigg(\dfrac{1}{2}-\dfrac{2}{3}\bigg)x=\dfrac{7}{12}\\ ⇔\dfrac{-1}{6}x=\dfrac{7}{12}\\ ⇔x=\dfrac{7}{12}:\dfrac{-1}{6}\\ ⇔x=\dfrac{-7}{2}\\ \text{Vậy x = $\dfrac{-7}{2}$}\\ 9)\ 25\%x+1\dfrac{2}{5}=\dfrac{-3}{5}+x\\ ⇔\dfrac{1}{4}x+\dfrac{7}{5}-x=\dfrac{-3}{5}\\ ⇔\dfrac{1}{4}x-x=\dfrac{-3}{5}-\dfrac{7}{5}\\ ⇔\dfrac{-3}{4}x=-2\\ ⇔x=(-2):\dfrac{-3}{4}\\ ⇔x=\dfrac{8}{3}\\ \text{Vậy x = $\dfrac{8}{3}$}\\ 10)\ \bigg|2x-\dfrac{1}{3}\bigg|-\dfrac{1}{2}=\dfrac{1}{6}\\ ⇔\bigg|2x-\dfrac{1}{3}\bigg|=\dfrac{1}{6}+\dfrac{1}{2}\\ ⇔\bigg|2x-\dfrac{1}{3}\bigg|=\dfrac{2}{3}\\ ⇔\left[ \begin{array}{l}2x-\dfrac{1}{3}=\dfrac{2}{3}\\2x-\dfrac{1}{3}=\dfrac{-2}{3}\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{-1}{6}\end{array} \right.\\ \text{Vậy x = $\dfrac{1}{2}$ hoặc x = $\dfrac{-1}{6}$}\\ 11)\ \bigg(\dfrac{1}{2}x-\dfrac{1}{3}\bigg)^2=\dfrac{16}{25}\\ ⇔\bigg(\dfrac{1}{2}x-\dfrac{1}{3}\bigg)^2=\bigg(±\dfrac{4}{5}\bigg)^2\\ ⇔\left[ \begin{array}{l}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{4}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-4}{5}\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=\dfrac{34}{15}\\x=\dfrac{-14}{15}\end{array} \right.\\ \text{Vậy x = $\dfrac{34}{15}$ hoặc x = $\dfrac{-14}{15}$}\\ 12)\ (3x-1)\bigg(\dfrac{-1}{2}x+5\bigg)=0\\ ⇔\left[ \begin{array}{l}3x-1=0\\\dfrac{-1}{2}x+5=0\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=10\end{array} \right.\\ \text{Vậy x = $\dfrac{1}{3}$ hoặc x = 10}\)
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