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Question

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

thám tử · Accepted Answer

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

$\Leftrightarrow3\dfrac{1}{2}x-3-\dfrac{2}{3}x+1-\dfrac{5}{6}x-2=-3$

$\Rightarrow3\dfrac{1}{2}x-\dfrac{2}{3}x-\dfrac{5}{6}x=-3+3-1+2$

$\Rightarrow2x=1$

$\Rightarrow x=\dfrac{1}{2}$

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

$\Leftrightarrow3\dfrac{1}{2}x-3-\dfrac{2}{3}x+1-\dfrac{5}{6}x-2=-3$

$\Rightarrow3\dfrac{1}{2}x-\dfrac{2}{3}x-\dfrac{5}{6}x=-3+3-1+2$

$\Rightarrow2x=1$

$\Rightarrow x=\dfrac{1}{2}$

Vậy.......

Nguyễn Hoàng Đào · Answer

$\Rightarrow3\dfrac{1}{2}x-3-\dfrac{2}{3}x+1+-\dfrac{5}{6}x-2=-3$

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

$\Rightarrow2x-4=-3$

​$\Rightarrow2x=-3+4=1$

$\Rightarrow x=\dfrac{1}{2}$

Vay $x=\dfrac{1}{2}$

Nhớ like nhéCâu trả lời: $\Rightarrow3\dfrac{1}{2}x-3-\dfrac{2}{3}x+1+-\dfrac{5}{6}x-2=-3$

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

$\Rightarrow2x-4=-3$

​$\Rightarrow2x=-3+4=1$

$\Rightarrow x=\dfrac{1}{2}$

Vay $x=\dfrac{1}{2}$

Nhớ like nhé

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