Question

[Duy Đức Anh Nguyễn]

Bài 2.1

Answer 1

[trương khoa]

\(x^2-2\left(m-1\right)x+2m-4=0\)

\(\Delta'=\left[-\left(m-1\right)\right]^2-\left(2m-4\right)\)

\(\Delta'=m^2-2m+1-2m+4\)

\(\Delta'=m^2-4m+5\)

\(\Delta'=\left(m-2\right)^2+1>0\)(với mọi m)

Vậy pt luôn có 2 nghiệm x₁ ,x₂ với mọi m

theo đinh lí Vi-ét ta có

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)=2m-2\\x_1x_2=2m-4\end{matrix}\right.\)

ta có \(x_1^2+x_2^2=3\)

⇔\(\left(x_1+x_2\right)^2-2x_1x_2=3\)

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

⇔\(4m^2-8m+4-4m+8=3\)

⇔\(4m^2-12m+9=0\)

\(\Delta'=\left(-6\right)^2-4\cdot9=0\)

⇒Pt có nghiệm kép

\(m_1=m_2=\dfrac{6}{4}=\dfrac{3}{2}\)

vậy m =\(\dfrac{3}{2}\) 

Answer 2

[missing you =]

\(x^2-2\left(m-1\right)x+2m-4=0\)(1)

\(\Delta'=\left[-\left(m-1\right)\right]^2-\left(2m-4\right)=m^2-2m+1-2m+4\)

\(=m^2-4m+5=\left(m-2\right)^2+1>0\left(\forall m\right)\)

\(=>\Delta'>0\) nên pt (1) luôn có 2 nghiệm phân biệt \(\forall m\\ \)

theo vi ét=>\(\left\{{}\begin{matrix}x1+x2=2\left(m-1\right)=2m-2\\x1x2=2m-4\end{matrix}\right.\)

có \(x1^2+x2^2=3< =>\left(x1+x2\right)^2-2x1x2-3=0\)

\(< =>\left(2m-2\right)^2-2\left(2m-4\right)-3=0\)

\(< =>4m^2-8m+4-4m+5=0\)

\(< =>4m^2-12m+9=0\)(2)

\(\Delta1=\left(-12\right)^2-4.9.4=0\)

=>pt (2) có nghiệm kép m1=m2=\(\dfrac{12}{2.4}=\dfrac{12}{8}=1,5\)

vậy m=1,5....