Question

Câu 8 á mọi người, giúp mình với!!!

Answer 1

Bài 8:

a: \(x^2-4x+m-2=0\)

\(\text{Δ}=\left(-4\right)^2-4\left(m-2\right)\)

\(=16-4m+8\)

=-4m+24

Để phương trình có nghiệm thì Δ>=0

=>-4m+24>=0

=>-4m>=-24

=>\(m< =6\)

b: Theo Vi-et, ta có:

\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4\); \(x_1x_2=\dfrac{c}{a}=\dfrac{m-2}{1}=m-2\)

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=4^2-2\left(m-2\right)\)

\(=16-2m+4=20-2m\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)

\(=4^2-4\left(m-2\right)\)

\(=16-4m+8=24-4m\)

\(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)

\(=4^3-3\cdot4\cdot\left(m-2\right)\)

\(=64-12m+24=88-12m\)