\(b,\)\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right).\left(\dfrac{x-1}{\sqrt{x}-1}-2\right)\)\(\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\dfrac{x-1-2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-1-2\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}^2-2\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
Vậy \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\) với \(x\ge0,x\ne1\)
\(c,x^2-6x+2m+3=0\)
Theo Vi-ét,ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=6\\x_1x_2=\dfrac{c}{a}=2m+3\end{matrix}\right.\)
Ta có :
\(\left(x_1-x_2\right)^2-2x_1x_2< 1\)
\(\Leftrightarrow x_1^2-2x_1x_2+x_2^2-2x_1x_2< 1\)
\(\Leftrightarrow x_1^2+x_2^2-4x_1x_2< 1\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-4x_1x_2< 1\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-6x_1x_2-1< 0\)
\(\Leftrightarrow6^2-6\left(2m+3\right)-1< 0\)
\(\Leftrightarrow-12m-18+35< 0\)
\(\Leftrightarrow-12m< -17\)
\(\Leftrightarrow m>\dfrac{17}{12}\)
