ĐKXĐ: \(x\ge\frac{1}{2}\)

\(\Leftrightarrow x^2-x-1-\frac{13x^2-28x+24}{2x+1}+x+1-2\sqrt{2x-1}=0\)

\(\Leftrightarrow\frac{2x^3-14x^2+25x-25}{2x+1}+\frac{\left(x+1\right)^2-4\left(2x-1\right)}{x+1+2\sqrt{2x-1}}=0\)

\(\Leftrightarrow\frac{\left(x-5\right)\left(2x^2-4x+5\right)}{2x+1}+\frac{\left(x-1\right)\left(x-5\right)}{x+1+2\sqrt{2x-1}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2x^2-4x+5}{2x+1}+\frac{x-1}{x+1+2\sqrt{2x-1}}\right)=0\)

\(\Rightarrow x=5\) (ngoặc to luôn dương với \(x\ge\frac{1}{2}\))

\(a)\)

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

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

\(\Leftrightarrow x^2-2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b)\)

\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)

\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)

\(\Leftrightarrow x^2+2x+1=x^2+10\)

\(\Leftrightarrow2x-9=0\)

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=\frac{2}{9}\)

đánh giá đi bạn 

\(\frac{6}{-x^2+10x-24}=\frac{6}{1-\left(x-5\right)^2}\ge6\)