x1 là nghiệm của phương trình
=>\(x_1^2-5x_1+3=0\)
=>\(x_1^2-4x_1+4-x_1-1=0\)
=>\(\left(x_1-2\right)^2=x_1+1\)
=>\(\sqrt{\left(x_1-2\right)^2}=\sqrt{x_1+1}\)
=>\(\left|x_1-2\right|=\sqrt{x_1+1}\)
x2 là nghiệm của phương trình
=>\(x_2^2-5x_2+3=0\)
=>\(x_2^2-4x_2+4-x_2-1=0\)
=>\(\left(x_2-2\right)^2=x_2+1\)
=>\(\sqrt{\left(x_2-2\right)^2}=\sqrt{x_2+1}\)
=>\(\left|x_2-2\right|=\sqrt{x_2+1}\)
Theo Vi-et, ta có:
\(\begin{cases}x_1+x_2=-\frac{b}{a}=5\\ x_1x_2=\frac{c}{a}=3\end{cases}\)
\(\left(x_1+1\right)\left(x_2+1\right)=x_1x_2+x_1+x_2+1\)
=5+3+1=9
\(A=\frac{x_1}{\left|x_1-2\right|}+\frac{x_2}{\left|x_2-2\right|}\)
\(=\frac{x_1}{\sqrt{x_1+1}}+\frac{x_2}{\sqrt{x_2+1}}=\frac{x_1+1-1}{\sqrt{x_1+1}}+\frac{x_2+1-1}{\sqrt{x_2+1}}\)
\(=\sqrt{x_1+1}+\sqrt{x_2+1}-\left(\frac{1}{\sqrt{x_1+1}}+\frac{1}{\sqrt{x_2+1}}\right)\)
Đặt \(B=\sqrt{x_1+1}+\sqrt{x_2+1}\)
=>\(B^2=x_1+1+x_2+1+2\sqrt{\left(x_1+1\right)\left(x_2+1\right)}\)
=>\(B^2=5+2+2\cdot\sqrt9=7+2\cdot3=13\)
=>\(B=\sqrt{13}\)
\(\frac{1}{\sqrt{x_1+1}}+\frac{1}{\sqrt{x_2+1}}=\frac{\sqrt{x_1+1}+\sqrt{x_2+1}}{\sqrt{\left(x_1+1\right)\left(x_2+1\right)}}\)
\(=\frac{\sqrt{13}}{\sqrt9}=\frac{\sqrt{13}}{3}\)
=>\(A=\sqrt{13}+\frac{\sqrt{13}}{3}=\frac{4\sqrt{13}}{3}\)
