Question

Cho pt: x^2-2(m+1)x+m-5=0

a) C/minh pt có 2 nghiệm pb x_1, ,x_{2 với mọi m}

b) tìm m để (X₁+1)²×X₂+(X₂+1)²×X₁=-16

Answer 1

a)Ta có:\(\Delta'=\left(m+1\right)^2-m+5=m^2+2m+1-m+5=m^2+m+6=\left(m+\frac{1}{2}\right)^2+\frac{23}{4}>0\)

=>pt luôn có 2 nghiệm pb với mọi m

b)Theo hệ thức vi-ét ta có:\(\left\{{}\begin{matrix}a+b=2m+2\\ab=m-5\end{matrix}\right.\)

\(\left(a+1\right)^2\cdot b+\left(b+1\right)^2a=-16\)

\(\Leftrightarrow\left(a^2+2a+1\right)b+\left(b^2+2b+1\right)a=-16\)

\(\Leftrightarrow a^2b+2ab+b+b^2a+2ab+a=-16\)

\(\Leftrightarrow ab\left(a+b\right)+4ab+a+b=-16\)

\(\Leftrightarrow\)\(\left(2m+2\right)\left(m-5\right)+4m-20+2m+2=-16\)

\(\Leftrightarrow2m^2-10m+2m-10+4m-20+2m+2+16=0\)

\(\Leftrightarrow2m^2-2m-12=0\)

\(\Leftrightarrow m^2-m-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=-2\end{matrix}\right.\)