Giải:
\(\dfrac{x-1}{2}+\dfrac{2-x}{3}\le\dfrac{3x-3}{4}\)
\(\Leftrightarrow\dfrac{6\left(x-1\right)}{12}+\dfrac{4\left(2-x\right)}{12}\le\dfrac{3\left(3x-3\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+4\left(2-x\right)\le3\left(3x-3\right)\)
\(\Leftrightarrow6x-6+8-4x\le9x-9\)
\(\Leftrightarrow2x+2\le9x-9\)
\(\Leftrightarrow2+9\le9x-2x\)
\(\Leftrightarrow11\le7x\)
\(\Leftrightarrow7x\ge11\)
\(\Leftrightarrow x\ge\dfrac{11}{7}\)
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