\(\frac{1}{x\left(x-1\right)}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{3}{4}\)(ĐKXĐ:\(x\ne0;1;-1;-2\))
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x+2\right)+x\left(x-1\right)}{\left(x-1\right)x\left(x+1\right)\left(x+2\right)}=\frac{3}{4}\)
\(\Leftrightarrow\frac{x^2+x+2x+2+x^2-x+2x-2+x^2-x}{\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]}=\frac{3}{4}\)
\(\Leftrightarrow\frac{3x^2+3x}{\left(x^2-x+2x-2\right)x\left(x+1\right)}=\frac{3}{4}\)
\(\Leftrightarrow\frac{3}{x^2+x-2}=\frac{3}{4}\)
=> x2 + x - 2 = 4
=> x2 + x - 6 = 0
=> \(\left(x+\frac{1}{2}\right)^2=\frac{25}{4}\)
Pt có nghiệm nhỏ nhất khi \(x+\frac{1}{2}=-\frac{5}{2}\)\(\Leftrightarrow x=-3\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}=\dfrac{1}{x-1}-\dfrac{1}{x+2}=\dfrac{x+2-x+1}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\\ \)
\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{4}\Rightarrow\left(x-1\right)\left(x+2\right)=4\)
\(\Rightarrow\left[\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
