Question

Cho x²-mx+1=0, tính:

a, x₁³+x₂³

b, \(\frac{x_1^2}{x_2^2}+\frac{x_2^2}{x_1^2}\)

Answer 1

a , \(x^3_1+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\) ( 1 )

Theo viet : \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=1\end{matrix}\right.\)

Thay vào ( 1 ) ta được : \(m^3-3.1.\left(m\right)=m^3-3m\)

Answer 2

Theo hệ thức vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=1\end{matrix}\right.\)

a)\(x_1^3+x_2^3=\left(x_1+x_2\right)\left(x_1^2-x_1x_2+x_2^2\right)=m\left[\left(x_1+x_2\right)^2-3x_1x_2\right]\)

\(=m\left(m^2-3\right)=m^3-3m\)

b)\(\frac{x_1^2}{x_2^2}+\frac{x_2^2}{x_1^2}=x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2=\left[\left(x_1+x_2\right)^2-2\right]^2-2\)

\(=\left(a^2-2\right)^2-2=a^4-4a^2+2\)