Question

Giải các phương trình

\(\frac{x}{2\left(x-3\right)}\) + \(\frac{x}{2\left(x+1\right)}\) = \(\frac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\frac{1}{x-1}\) + \(\frac{2}{x^2+x+1}\) = \(\frac{3x^2}{x^3-1}\)

Ai rảnh thì kết bạn với mik nha!!!

Answer 1

\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\)

=> 2x=0

<=> x=0

Vậy x=0

Answer 2

+ Ta có: \(\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}=\frac{2x}{\left(x+1\right).\left(x-3\right)}\)\(\left(ĐKXĐ: x\ne-1, x\ne3\right)\)

      \(\Leftrightarrow\frac{x.\left(x+1\right)+x.\left(x-3\right)}{2.\left(x-3\right).\left(x+1\right)}=\frac{4x}{2.\left(x-3\right).\left(x+1\right)}\)

       \(\Rightarrow x^2+x+x^2-3x=4x\)

      \(\Leftrightarrow\left(x^2+x^2\right)+\left(x-3x-4x\right)=0\)

      \(\Leftrightarrow2x^2-6x=0\)

      \(\Leftrightarrow2x.\left(x-6\right)=0\)

      \(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=6\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{0,6\right\}\)

+ Ta có: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)\(\left(ĐKXĐ:x\ne1,x^2+x+1\ne0\right)\)

       \(\Leftrightarrow\frac{\left(x^2+x+1\right)+2.\left(x-1\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right).\left(x^2+x+1\right)}\)

        \(\Rightarrow x^2+x+1+2x-2=3x^2\)

      \(\Leftrightarrow\left(x^2-3x^2\right)+\left(x+2x\right)+\left(1-2\right)=0\)

      \(\Leftrightarrow-2x^2+3x-1=0\)

      \(\Leftrightarrow2x^2-3x+1=0\)

      \(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)

      \(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)=0\)

      \(\Leftrightarrow\left(2x-1\right).\left(x-1\right)=0\)

      \(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=1\left(L\right)\end{cases}}\)

Vậy \(S=\left\{\frac{1}{2}\right\}\)