Câu hỏi của Thuy Tran

Question

Tìm x:

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

Mysterious Person · Accepted Answer

ta có : $\dfrac{3}{2}\left(x-\dfrac{1}{5}\right)-x\left(x-\dfrac{3}{5}\right)=-x^2$

$\Leftrightarrow\dfrac{3}{2}x-\dfrac{3}{10}-x^2+\dfrac{3}{5}x=-x^2$ $\Leftrightarrow\dfrac{21}{10}x-\dfrac{3}{10}=0$

$\Leftrightarrow x=\dfrac{1}{7}$ vậy $x=\dfrac{1}{7}$Câu trả lời: ta có : $\dfrac{3}{2}\left(x-\dfrac{1}{5}\right)-x\left(x-\dfrac{3}{5}\right)=-x^2$

$\Leftrightarrow\dfrac{3}{2}x-\dfrac{3}{10}-x^2+\dfrac{3}{5}x=-x^2$ $\Leftrightarrow\dfrac{21}{10}x-\dfrac{3}{10}=0$

$\Leftrightarrow x=\dfrac{1}{7}$ vậy $x=\dfrac{1}{7}$

reina mikichi · Answer

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

$\dfrac{3}{2}\cdot x-\dfrac{3}{2}\cdot\dfrac{1}{5}-x^2-x\cdot\dfrac{3}{5}=-x^2$

$x\left(\dfrac{3}{2}-\dfrac{3}{5}\right)-\dfrac{3}{10}-x^2+x^2=0$

$x\cdot\dfrac{9}{10}=\dfrac{3}{10}$

x=$\dfrac{3}{10}:\dfrac{9}{10}$

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

$\dfrac{3}{2}\cdot x-\dfrac{3}{2}\cdot\dfrac{1}{5}-x^2-x\cdot\dfrac{3}{5}=-x^2$

$x\left(\dfrac{3}{2}-\dfrac{3}{5}\right)-\dfrac{3}{10}-x^2+x^2=0$

$x\cdot\dfrac{9}{10}=\dfrac{3}{10}$

x=$\dfrac{3}{10}:\dfrac{9}{10}$

x=$\dfrac{1}{3}$

Violympic toán 7