Question

[Hoàng Nguyệt]

Câu 1 giúp e với , cảm ơn trước

Answer 1

[Nguyễn Lê Phước Thịnh]

Ta có: \(\Delta=7^2-4\cdot2\cdot\left(-1\right)=49+9=58>0\)

nên phương trình có hai nghiệm phân biệt

a) Áp dụng hệ thức Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1\cdot x_2=\dfrac{-1}{2}\\x_1+x_2=\dfrac{-7}{2}\end{matrix}\right.\)

Ta có: \(\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}\)

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

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

\(=\dfrac{\left(-\dfrac{7}{2}\right)^2-2\cdot\dfrac{-1}{2}}{\left(-\dfrac{1}{2}\right)^2}=\dfrac{\dfrac{49}{4}+1}{\dfrac{1}{4}}=\dfrac{53}{4}\cdot4=53\)

Answer 2

[Nguyễn Việt Lâm]

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{7}{2}\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{1}{x_1^3}+\dfrac{1}{x_2^3}=\dfrac{x_1^3+x_2^3}{\left(x_1x_2\right)^3}=\dfrac{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}{\left(x_1x_2\right)^3}=\dfrac{\left(-\dfrac{7}{2}\right)^3-3.\left(-\dfrac{1}{2}\right).\left(-\dfrac{7}{2}\right)}{\left(-\dfrac{1}{2}\right)^3}=385\)

\(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\left(-\dfrac{7}{2}\right)^3-3\left(-\dfrac{1}{2}\right).\left(-\dfrac{7}{2}\right)=-\dfrac{385}{8}\)