Question

Cho: \(5x^2+7x-1=0\)
Tính: \(A=\left(x_1+\dfrac{7}{5}\right)x_1+\dfrac{1}{25x_2^2}+x_2^2\)

Answer 1

\(x_1\) là nghiệm nên:

\(5x_1^2+7x_1-1=0\)

\(\Leftrightarrow5\left(x_1^2+\dfrac{7}{5}x_1\right)=1\)

\(\Leftrightarrow\left(x_1+\dfrac{7}{5}\right)x_1=\dfrac{1}{5}\)

\(\Rightarrow A=\dfrac{1}{5}+\dfrac{1}{25x_2^2}+x_2^2=\left(x_2-\dfrac{1}{5x_2}\right)^2+\dfrac{3}{5}\)

Do \(x_2\) là nghiệm nên:

\(5x_2^2+7x_2-1=0\)

\(\Leftrightarrow5x_2^2-1=7x_2\)

\(\Leftrightarrow\dfrac{5x_2^2-1}{5x_2}=\dfrac{7}{5}\)

\(\Leftrightarrow x_2-\dfrac{1}{5x_2}=\dfrac{7}{5}\)

\(\Rightarrow A=\left(\dfrac{7}{5}\right)^2+\dfrac{3}{5}=\dfrac{64}{25}\)