Question

[Lương Thị Ngọc Hải]

                    Tìm x biết

     \(\dfrac{3}{x}+\dfrac{x}{x+1}+\dfrac{x-3}{x}=\dfrac{13}{7}\)

Answer 1

[Hquynh]

\(đk:x\ne0;x\ne-1\\ \dfrac{3}{x}+\dfrac{x}{x+1}+\dfrac{x-3}{x}=\dfrac{13}{7}\\ \Leftrightarrow\dfrac{7.3.\left(x+1\right)}{7x\left(x+1\right)}+\dfrac{x.x.7}{7x\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right).7}{7x\left(x+1\right)}=\dfrac{13.x.\left(x+1\right)}{7x\left(x+1\right)}\\ \Leftrightarrow\dfrac{21x+21+7x^2+7x^2+7x-21x-21}{7x\left(x+1\right)}=\dfrac{13x^2+13x}{7x\left(x+1\right)}\\ \Leftrightarrow14x^2+7x=13x^2+13x\\ \Leftrightarrow14x^2-13x^2=13x-7x\\ \Leftrightarrow x^2=6x\\ \Leftrightarrow x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=6\left(t/m\right)\end{matrix}\right.\Rightarrow x=6\)