Question

[mon an]

Giải hết nha

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(x^2+2\left(m+1\right)x-m^2-3=0\)

Thay m=0 vào phương trình, ta được:

\(x^2+2\left(0+1\right)x-0^2-3=0\)

=>\(x^2+2x-3=0\)

=>(x+3)(x-1)=0

=>\(\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

b: Vì \(a\cdot c=-m^2-3< =-3< 0\forall m\)

nên phương trình luôn có hai nghiệm phân biệt

c: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\left(m+1\right)\\x_1x_2=\dfrac{c}{a}=-m^2-3\end{matrix}\right.\)

\(x_1^2+x_2^2+x_1x_2=20\)

=>\(\left(x_1+x_2\right)^2-x_1x_2=20\)

=>\(\left(-2m-2\right)^2-\left(-m^2-3\right)=20\)

=>\(4m^2+8m+4+m^2+3=20\)

=>\(5m^2+8m-13=0\)

=>\(\left(5m+13\right)\left(m-1\right)=0\)

=>\(\left[{}\begin{matrix}m=-\dfrac{13}{5}\\m=1\end{matrix}\right.\)

d: \(\left(2x_1+1\right)\left(2x_2+1\right)+15=0\)

=>\(4x_1x_2+2\left(x_1+x_2\right)+16=0\)

=>\(2x_1x_2+\left(x_1+x_2\right)+8=0\)

=>\(2\left(-m^2-3\right)-2\left(m+1\right)+8=0\)

=>\(-2m^2-6-2m-2+8=0\)

=>\(2m^2+2m=0\)

=>2m(m+1)=0

=>\(\left[{}\begin{matrix}m=0\\m=-1\end{matrix}\right.\)

e: \(P=x_1^2+x_2^2+5x_1x_2\)

\(=\left(x_1+x_2\right)^2+3x_1x_2\)

\(=\left[-2\left(m+1\right)\right]^2+3\left(-m^2-3\right)\)

\(=4m^2+8m+4-3m^2-9\)

\(=m^2+8m-5\)

\(=m^2+8m+16-21=\left(m+4\right)^2-21>=-21\forall m\)

Dấu '=' xảy ra khi m+4=0

=>m=-4