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Question

$\dfrac{-2}{3}x+\dfrac{-3}{7}+\dfrac{1}{2}x=\dfrac{-5}{6}$

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thám tử · Accepted Answer

$\dfrac{-2}{3}x+\dfrac{-3}{7}+\dfrac{1}{2}x=\dfrac{-5}{6}$

$\Rightarrow\dfrac{-2}{3}x+\dfrac{1}{2}x=\dfrac{-5}{6}-\left(\dfrac{-3}{7}\right)$

$\Rightarrow\dfrac{-1}{6}x=\dfrac{-17}{42}$

$\Rightarrow x=\dfrac{-17}{42}:\left(\dfrac{-1}{6}\right)$

$\Rightarrow x=\dfrac{17}{7}$

Vậy...Câu trả lời: $\dfrac{-2}{3}x+\dfrac{-3}{7}+\dfrac{1}{2}x=\dfrac{-5}{6}$

$\Rightarrow\dfrac{-2}{3}x+\dfrac{1}{2}x=\dfrac{-5}{6}-\left(\dfrac{-3}{7}\right)$

$\Rightarrow\dfrac{-1}{6}x=\dfrac{-17}{42}$

$\Rightarrow x=\dfrac{-17}{42}:\left(\dfrac{-1}{6}\right)$

$\Rightarrow x=\dfrac{17}{7}$

Vậy...

Hải Đăng · Answer

$\dfrac{-2}{3}x+\dfrac{-3}{7}+\dfrac{1}{2}x=\dfrac{5}{6}$

$\dfrac{-2}{3}x+\dfrac{1}{2}x=\dfrac{-5}{6}-\dfrac{-3}{7}$

$\dfrac{-2}{3}x+\dfrac{1}{2}x=\dfrac{-17}{42}$

$\left(\dfrac{-2}{3}+\dfrac{1}{2}\right)x=\dfrac{-17}{42}$

$\dfrac{-1}{6}x=\dfrac{-17}{42}$

$x=\dfrac{-17}{42}:\dfrac{1}{6}$

$x=\dfrac{-17}{7}$

Vậy .....

Chúc bạn học tốt!Câu trả lời: $\dfrac{-2}{3}x+\dfrac{-3}{7}+\dfrac{1}{2}x=\dfrac{5}{6}$