Question

x1+x2=5; x1x2=3

tính \(\dfrac{1}{5x_1-2}+\dfrac{1}{x_2^2+1}\)

Answer 1

\(x_1+x_2=5;x_1\cdot x_2=3\)

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2\cdot x_1x_2\)

\(=5^2-2\cdot3=25-6=19\)

\(\dfrac{1}{5x_1-2}+\dfrac{1}{x_2^2+1}\)

\(=\dfrac{1}{x_1\left(x_1+x_2\right)-2}+\dfrac{1}{x_2^2+1}\)

\(=\dfrac{1}{x_1^2+x_1\cdot x_2-2}+\dfrac{1}{x_2^2+1}\)

\(=\dfrac{1}{x_1^2+3-2}+\dfrac{1}{x_2^2+1}\)

\(=\dfrac{1}{x_1^2+1}+\dfrac{1}{x_2^2+1}\)

\(=\dfrac{x_2^2+1+x_1^2+1}{\left(x_1^2+1\right)\left(x_2^2+1\right)}\)

\(=\dfrac{19+2}{\left(x_1x_2\right)^2+\left(x_1^2+x_2^2+1\right)}=\dfrac{21}{3^2+\left(19+1\right)}=\dfrac{21}{9+20}=\dfrac{21}{29}\)