Question

  cho pt 2x²-3x-4=0 có hai nghiệm x₁ x₂ , không giải pt  hãy tính giá trị biểu thức : \(\dfrac{x_1^2}{x_1-2}\) + \(\dfrac{x_{\text{2}}^2}{x_{\text{2}}-2}\)

Answer 1

Theo hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-2\end{matrix}\right.\)

Lại có: \(\dfrac{x_1^2}{x_1-2}+\dfrac{x_2^2}{x_2-2}=\dfrac{x_1^2}{x_1+x_1x_2}+\dfrac{x_2^2}{x_2+x_1x_2}\)

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

\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+\left(x_1+x_2\right)}{x_1x_2+\left(x_1+x_2\right)+1}\)

\(=\dfrac{\left(\dfrac{3}{2}\right)^2-2\cdot\left(-2\right)+\dfrac{3}{2}}{-2+\dfrac{3}{2}+1}=\dfrac{31}{2}\)

$\text{#}Toru$