ta có : \(\dfrac{2x-5}{4x+1}=\dfrac{3x-7}{6x-11}\Leftrightarrow\left(2x-5\right)\left(6x-11\right)=\left(3x-7\right)\left(4x+1\right)\)
\(\Leftrightarrow12x^2-22x-30x+55=12x^2+3x-28x-7\)
\(\Leftrightarrow12x^2-22x-30x+55-12x^2-3x+28x+7=0\)
\(\Leftrightarrow-27x+62=0\Leftrightarrow27x=62\Leftrightarrow x=\dfrac{62}{27}\) vậy \(x=\dfrac{62}{27}\)
Đúng 0
Bình luận (0)
\(\dfrac{2x-5}{4x+1}=\dfrac{3x-7}{6x-11}\)
\(\Rightarrow\left(2x-5\right)\left(6x-11\right)=\left(3x-7\right)\left(4x+1\right)\)
\(\Rightarrow12x^2-52x+55=12x^2-25x-7\)
\(\Rightarrow52x-55=25x+7\)
\(\Rightarrow27x=62\)
\(\Rightarrow x=\dfrac{62}{27}\)
Vậy...
Đúng 0
Bình luận (0)
