Question

Cho phương trình: \(3x^2+4x-1=0_{ }\) có 2 nghiệm phân biệt x_1,x2. Không giải phương trình hãy tính giá trị biểu thức: \(A=\dfrac{x^3_1x^2_2\left(1-3x^2_2\right)}{x^2_1+x^2_2}\)

Answer 1

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-\dfrac{4}{3}\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)

\(A=\dfrac{x_1^3\cdot x_2^2\left(1-3x_2^2\right)}{x_1^2+x_2^2}\)

\(=\dfrac{\left(x_1x_2\right)^2\cdot\left(x_1-3x_1\cdot x_2^2\right)}{\left(x_1+x_2\right)^2-2x_1x_2}\)

\(=\dfrac{\left(-\dfrac{1}{3}\right)^2\left(x_1-3\cdot x_2\cdot\dfrac{-1}{3}\right)}{\left(-\dfrac{4}{3}\right)^2-2\cdot\dfrac{-1}{3}}\)

\(=\dfrac{\dfrac{1}{9}\left(x_1+x_2\right)}{\dfrac{16}{9}+\dfrac{2}{3}}=\dfrac{\dfrac{1}{9}\cdot\dfrac{-4}{3}}{\dfrac{22}{9}}=\dfrac{-4}{27}:\dfrac{22}{9}=-\dfrac{2}{33}\)