1,Giải sử x₀ là nghiệm chung của hai pt

Ta có hệ: \(\left\{{}\begin{matrix}x_0^2-\left(m+2\right)x_0+3m-1=0\left(1\right)\\x_0^2-\left(2m+3\right)x_0+3m+3=0\end{matrix}\right.\)

=> \(\left(2m+3\right)x_0-\left(m+2\right)x_0+3m-1-3m-3=0\)

<=> \(x_0\left(m+1\right)-4=0\)

Do hai pt có nghiệm chung => \(x_0\in R\) => \(m\ne-1\)

<=> \(x_0=\frac{4}{m+1}\) thay vào (1) có

\(\frac{16}{\left(m+1\right)^2}-\frac{\left(m+2\right).4}{m+1}+3m-1=0\)

<=> \(\frac{16}{\left(m+1\right)^2}-\frac{4\left(m+2\right)\left(m+1\right)}{\left(m+1\right)^2}+\frac{3m\left(m+1\right)^2}{\left(m+1\right)^2}-\frac{\left(m+1\right)^2}{\left(m+1\right)^2}=0\)

<=> \(16-4\left(m^2+3m+2\right)+3m\left(m^2+2m+1\right)-\left(m^2+2m+1\right)=0\)

<=> \(16-4m^2-12m-8+3m^3+6m^2+3m-m^2-2m-1=0\)

<=> \(3m^3+m^2-11m+7=0\)

<=> \(3m^3-3m^2+4m^2-4m-7m+7=0\)

<=>\(3m^2\left(m-1\right)+4m\left(m-1\right)-7\left(m-1\right)=0\)

<=> \(\left(m-1\right)\left(3m^2+4m-7\right)=0\)

<=> \(\left(m-1\right)^2\left(3m+7\right)=0\)

<=> \(\left[{}\begin{matrix}m=1\\m=-\frac{7}{3}\end{matrix}\right.\)

@@ cái gì vậy!!

Lời giải:

a)

\(f(-3)=(-3)^2=9; f(-\frac{1}{2})=(\frac{-1}{2})^2=\frac{1}{4}\)

\(f(0)=0^2=0\)

\(g(1)=3-1=2; g(2)=3-2=1; g(3)=3-3=0\)

b)

\(2f(a)=g(a)\)

\(\Leftrightarrow 2a^2=3-a\)

\(\Leftrightarrow 2a^2+a-3=0\Leftrightarrow (2a+3)(a-1)=0\)

\(\Rightarrow \left[\begin{matrix} a=\frac{-3}{2}\\ a=1\end{matrix}\right.\)

\(f\left(x\right)=\left|x-1\right|+\left|4-x\right|+2\left(\left|x-2\right|+\left|4-x\right|\right)+\left|x-3\right|+\left|4-x\right|+2\left|x-3\right|\)

\(f\left(x\right)\ge\left|x-1+4-x\right|+2\left|x-2+4-x\right|+\left|x-3+4-x\right|+2\left|x-3\right|\)

\(f\left(x\right)\ge3+4+1+2\left|x-3\right|=8+2\left|x-3\right|\ge8\)

\(\Rightarrow f\left(x\right)_{min}=8\) khi \(x=3\)

\(f\left(\left|a\right|\right)=2\Leftrightarrow\left(\sqrt{3}+1\right)\left|a\right|-2\sqrt{3}=2\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)\left|a\right|=2\sqrt{3}+2=2\left(\sqrt{3}+1\right)\)

\(\Leftrightarrow\left|a\right|=2\Rightarrow a=\pm2\)

\(f\left(x\right)⋮\left(x-1\right)\left(x+2\right)\Leftrightarrow\left\{{}\begin{matrix}f\left(1\right)=0\\f\left(-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+\left(a+b\right)+\left(2+b\right)+1=0\\-8a+4\left(a+b\right)-2\left(2+b\right)+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a+2b=-3\\-4a+2b=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=-\frac{1}{2}\end{matrix}\right.\)