\(\dfrac{10+x}{17+x}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(10+x\right)=3\left(17+x\right)\)
\(40+4x=51+3x\)
\(4x-3x=51-40\)
\(x=11\)
Vậy....
\(\dfrac{40+x}{77-x}=\dfrac{6}{7}\)
\(\Leftrightarrow7\left(40+x\right)=6\left(77-x\right)\)
\(280+7x=462-6x\)
\(462-280+7x=6x\)
\(182+7x=6x\)
\(182=-1x\)
\(x=-182\)