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a)
\(f\left(x\right)=3x^2-5x+1\)
\(3f\left(x\right)=9x^2-15x+3\)
\(3f\left(x\right)=\left(9x^2-15x+\frac{25}{4}\right)-\frac{13}{4}\)
\(3f\left(x\right)=\left(3x-\frac{5}{2}\right)^2-\frac{13}{4}\)
Mà \(\left(3x-\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3f\left(x\right)\ge\frac{-13}{4}\)
\(\Leftrightarrow f\left(x\right)\ge-\frac{13}{12}\)
Dấu '=' xảy ra khi :
\(3x-\frac{5}{2}=0\Leftrightarrow3x=\frac{5}{2}\Leftrightarrow x=\frac{5}{6}\)
\(f\left(x\right)=2x^2-9x-3\)
\(2f\left(x\right)=4x^2-18x-6\)
\(2f\left(x\right)=\left(4x^2-18x+\frac{81}{4}\right)-\frac{105}{4}\)
\(2f\left(x\right)=\left(2x-\frac{9}{2}\right)^2-\frac{105}{4}\)
Mà \(\left(2x-\frac{9}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2f\left(x\right)\ge-\frac{105}{4}\)
\(\Leftrightarrow f\left(x\right)\ge-\frac{105}{8}\)
Dấu "=" xảy ra khi :
\(2x-\frac{9}{2}=0\Leftrightarrow2x=\frac{9}{2}\Leftrightarrow x=\frac{9}{4}\)
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
1) \(A=\frac{2018x^2-2.2018x+2018^2}{2018x^2}=\frac{\left(x-2018\right)^2+2017x^2}{2018x^2}=\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\)
vì \(\frac{\left(x-2018\right)^2}{2018x^2}\ge0\Rightarrow\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\ge\frac{2017}{2018}\)
dấu = xảy ra khi x-2018=0
=> x=2018
Vậy Min A=\(\frac{2017}{2017}\)khi x=2018
2) \(B=\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}=1+\frac{10}{3.x^2+9x+7}\)
\(=1+\frac{10}{3.\left(x^2+9x\right)+7}=1+\frac{10}{3.\left[x^2+\frac{2.x.3}{2}+\left(\frac{3}{2}\right)^2\right]-\frac{9}{4}+7}=1+\frac{10}{3.\left(x+\frac{9}{2}\right)^2+\frac{1}{4}}\)
để B lớn nhất => \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\)nhỏ nhất
mà \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)vì \(3.\left(x+\frac{3}{2}\right)^2\ge0\)
dấu = xảy ra khi \(x+\frac{3}{2}=0\)
=> x=\(-\frac{3}{2}\)
Vậy maxB=\(41\)khi x=\(-\frac{3}{2}\)
3) \(M=\frac{3x^2+14}{x^2+4}=\frac{3.\left(x^2+4\right)+2}{x^2+4}=3+\frac{2}{x^2+4}\)
để M lớn nhất => x2+4 nhỏ nhất
mà \(x^2+4\ge4\)(vì x2 lớn hơn hoặc bằng 0)
dấu = xảy ra khi x2 =0
=> x=0
Vậy Max M\(=\frac{7}{2}\)khi x=0
ps: bài này khá dài, sai sót bỏ qua =))