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\(\Delta'=b'^2-ac=-6m+7=>\)\(m\ge\frac{7}{6}\)
Theo Vi-ét : \(\hept{\begin{cases}x_1+x_2=2\left(m-2\right)\\x_1.x_2=m^2+2m-3\end{cases}}\)Mà \(\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{5}=>\)\(\frac{x_1+x_2}{x_1.x_2}=\frac{x_1+x_2}{5}\)
=> \(x_1.x_2=5\)<=> \(m^2+2m-3=5\)<=> \(m^2+2m-8=0\)
Giải pt trên ta đc : \(\orbr{\begin{cases}m=2\\m=-4\end{cases}}\)Mà \(m\ge\frac{7}{6}\)=> \(m=2\)
Theo định lí Vi-et , ta có : \(\begin{cases}x_1+x_2=1\\x_1.x_2=-5\end{cases}\)
- \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=1-2.\left(-5\right)=11\)
- \(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=1-3.\left(-5\right).1=16\)
- \(C=\left(2x_1+x_2\right)\left(2x_2+x_1\right)=\left(1+x_1\right)\left(1+x_2\right)=\left(x_1+x_2\right)+x_1.x_2+1=1-5+1=-3\)
Xét \(\Delta=\left(2m-1\right)^2-8\left(m-1\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0\)
=> PT luôn có 2 nghiệm x1,x2 với mọi m
Theo hệ thức Viet ta có \(\hept{\begin{cases}x_1+x_2=\frac{1-2m}{2}\\x_1x_2=\frac{m-1}{2}\end{cases}}\)
\(\Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2=\left(\frac{1-2m}{2}\right)^2-\frac{3\left(m-1\right)}{2}\)
\(=\frac{1-4m+4m^2-6m+6}{4}=\frac{4m^2-10m+7}{4}\)
\(=\frac{\left(2m-\frac{5}{2}\right)^2+\frac{3}{4}}{4}\ge\frac{3}{16}\)
Dấu "=" xảy ra khi \(2m=\frac{5}{2}\Rightarrow m=\frac{5}{4}\Rightarrow\frac{a}{b}=\frac{5}{4}\)
\(\Rightarrow4a=5b\Rightarrow2a=\frac{5b}{2}\)
lúc đó \(P=\frac{5b}{2}+2b=\frac{9b}{2}\)
\(a,\Delta=m^2-4m+4=\left(m-2\right)^2\ge0\forall m\)
Nên pt đã cho luôn có 2 nghiệm phân biệt với mọi m
b, Theo Vi-ét \(\hept{\begin{cases}x_1+x_2=m\\x_1x_2=m-1\end{cases}}\)
Ta có \(B=\frac{2x_1x_2+3}{x_1^2+x_2^2+2\left(1+x_1x_2\right)}=1\)
\(\Leftrightarrow\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=1\)
\(\Leftrightarrow\frac{2\left(m-1\right)+3}{m^2+2}=1\)
\(\Leftrightarrow\frac{2m+1}{m^2+2}=1\)
\(\Leftrightarrow2m+1=m^2+2\)
\(\Leftrightarrow m^2-2m+1=0\)
\(\Leftrightarrow\left(m-1\right)^2=0\)
\(\Leftrightarrow m=1\)
Theo Vi-et, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-5\right)}{-3}=\dfrac{5}{-3}=-\dfrac{5}{3}\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{3}{-3}=-1\end{matrix}\right.\)
\(P=\dfrac{x_1-3}{x_1^2}+\dfrac{x_2-3}{x_2^2}\)
\(=\dfrac{x_1\cdot x_2^2-3\cdot x_2^2+x_2\cdot x_1^2-3\cdot x_1^2}{\left(x_1x_2\right)^2}=\dfrac{x_1x_2\left(x_1+x_2\right)-3\left(x_1^2+x_2^2\right)}{\left(x_1x_2\right)^2}\)
\(=\dfrac{-1\cdot\dfrac{-5}{3}-3\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{\left(-1\right)^2}\)
\(=\dfrac{5}{3}-3\cdot\left[\left(-\dfrac{5}{3}\right)^2-2\cdot\left(-1\right)\right]\)
\(=\dfrac{5}{3}-3\cdot\left[\dfrac{25}{9}+2\right]=\dfrac{5}{3}-3\cdot\dfrac{43}{9}=\dfrac{5}{3}-\dfrac{43}{3}=\dfrac{-38}{3}\)