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a: \(\Leftrightarrow\left(2m+1\right)^2-4\left(m^2-3\right)=0\)
\(\Leftrightarrow4m^2+4m+1-4m^2+12=0\)
=>4m=-13
hay m=-13/4
c: \(\Leftrightarrow\left(2m-2\right)^2-4m^2>=0\)
\(\Leftrightarrow4m^2-8m+4-4m^2>=0\)
=>-8m>=-4
hay m<=1/2
Δ=(2m-1)^2-4*2*(m-1)
=4m^2-4m+1-8m+8
=4m^2-12m+9=(2m-3)^2>=0
=>PT luôn có 2 nghiệm
4x1^2+4x2^2+2x1x2=0
=>4[(x1+x2)^2-2x1x2]+m-1=0
=>4[(-2m+1)^2/4-2*(m-1)/2]+m-1=0
=>(2m-1)^2-4(m-1)+m-1=0
=>4m^2-4m+1-3m+3=0
=>4m^2-7m+4=0
=>\(m\in\varnothing\)
a, Thay m=2 vào pt ta có:
(1)\(\Leftrightarrow2x^2+\left(2.2-1\right)x+2-1=0\)
\(\Leftrightarrow2x^2+3x+1=0\\ \Leftrightarrow\left(2x^2+2x\right)+\left(x+1\right)=0\\ \Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=-1\end{matrix}\right.\)
b,\(\Delta=\left(2m-1\right)^2-4.2\left(m-1\right)=4m^2-4m+1-8\left(m-1\right)=4m^2-4m+1-8m+8=4m^2-12m+9\)
Để pt có 2 nghiệm thì \(\Delta\ge0\Leftrightarrow4m^2-12m+9\ge0\left(luôn.đúng\right)\)
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1-2m}{2}\\x_1x_2=\dfrac{m-1}{2}\end{matrix}\right.\)
\(4x^2_1+4x^2_2+2x_1x_2=1\\ \Leftrightarrow4\left(x^2_1+x^2_2\right)+2.\dfrac{m-1}{2}=1\\ \Leftrightarrow4\left(x_1+x_2\right)^2-8x_1x_2+m-1=1\\ \Leftrightarrow4.\left(\dfrac{1-2m}{2}\right)^2-8.\dfrac{m-1}{2}+m-2=0\)
\(4.\dfrac{\left(1-2m\right)^2}{4}-4\left(m-1\right)+m-2=0\\ \Leftrightarrow4\left(1-4m+4m^2\right)-4m+4+m-2=0\\ \Leftrightarrow4-16m+16m^2-3m+2=0\\ \Leftrightarrow16m^2-19m+6=0\)
Ta có:\(\Delta=\left(-19\right)^2-4.16.6=361-384=-23< 0\)
Suy ra pt vô nghiệm
