Câu 7:
\(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot m^2\)
\(=\left(2m-2\right)^2-4m^2=\left(2m-2-2m\right)\left(2m-2+2m\right)\)
\(=-2\left(4m-2\right)=-8m+4\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>-8m+4>0
=>-8m>-4
=>\(m< \dfrac{1}{2}\)
Để phương trình vô nghiệm thì Δ<0
=>-8m+4<0
=>-8m<-4
=>\(m>\dfrac{1}{2}\)
Để phương trình có nghiệm kép thì Δ=0
=>-8m+4=0
=>\(m=\dfrac{1}{2}\)
Khi m=1/2 thì phương trình sẽ trở thành:
\(x^2-2\left(\dfrac{1}{2}-1\right)x+\left(\dfrac{1}{2}\right)^2=0\)
=>\(x^2+x+\dfrac{1}{4}=0\)
=>\(\left(x+\dfrac{1}{2}\right)^2=0\)
=>\(x+\dfrac{1}{2}=0\)
=>\(x=-\dfrac{1}{2}\)
Câu 9:
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\\x_1x_2=\dfrac{c}{a}=-3\end{matrix}\right.\)
\(P=x_1^3+x_2^3+2x_1x_2\)
\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+2x_1x_2\)
\(=\left(-2\right)^3-3\cdot\left(-2\right)\cdot\left(-3\right)+2\cdot\left(-3\right)\)
\(=-8-3\cdot6-6=-14-18=-32\)
\(Q=x_1^2+x_2^2-6x_1x_2\)
\(=\left(x_1+x_2\right)^2-8x_1x_2\)
\(=\left(-2\right)^2-8\cdot\left(-3\right)=4+24=28\)
\(T=\left(x_1-x_2\right)^2-6x_1x_2\)
\(=\left(x_1+x_2\right)^2-4x_1x_2-6x_1x_2\)
\(=\left(x_1+x_2\right)^2-10x_1x_2\)
\(=\left(-2\right)^2-10\cdot\left(-3\right)=30+4=34\)
\(K=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-2}{-3}=\dfrac{2}{3}\)
\(H=x_1^2\cdot x_2+x_2^2\cdot x_1=x_1x_2\left(x_1+x_2\right)\)
\(=\left(-2\right)\cdot\left(-3\right)=6\)
\(M=x_1^3\cdot x_2+x_1\cdot x_2^3\)
\(=x_1x_2\left(x_1^2+x_2^2\right)\)
\(=\left(-3\right)\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\)
\(=\left(-3\right)\left[\left(-2\right)^2-2\cdot\left(-3\right)\right]\)
\(=-3\cdot\left(4+6\right)=-30\)
