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Theo Cô si 4x+\frac{1}{4x}\ge24x+4x1≥2 , đẳng thức xảy ra khi và chỉ khi 4x=\frac{1}{4x}=1\Leftrightarrow x=\frac{1}{4}4x=4x1=1⇔x=41). Do đó
A\ge2-\frac{4\sqrt{x}+3}{x+1}+2016A≥2−x+14x+3+2016
A\ge4-\frac{4\sqrt{x}+3}{x+1}+2014A≥4−x+14x+3+2014
A\ge\frac{4x-4\sqrt{x}+1}{x+1}+2014=\frac{\left(2\sqrt{x}-1\right)^2}{x+1}+2014\ge2014A≥x+14x−4x+1+2014=x+1(2x−1)2+2014≥2014
Hơn nữa A=2014A=2014 khi và chỉ khi \left\{{}\begin{matrix}x=\dfrac{1}{4}\\2\sqrt{x}-1=0\end{matrix}\right.{x=412x−1=0 \Leftrightarrow x=\dfrac{1}{4}⇔x=41 .
Vậy GTNN = 2014
\(a,\Leftrightarrow y=0;x=2\Leftrightarrow2m-2+m-2=0\Leftrightarrow m=\dfrac{4}{3}\)
\(b,\) PT giao Ox: \(\Leftrightarrow\left(m-1\right)x=2-m\Leftrightarrow x=\dfrac{2-m}{m-1}\Leftrightarrow A\left(\dfrac{2-m}{m-1};0\right)\Leftrightarrow OA=\left|\dfrac{2-m}{m-1}\right|\)
PT giao Oy: \(y=m-2\Leftrightarrow B\left(0;m-2\right)\Leftrightarrow OB=\left|m-2\right|\)
\(S_{OAB}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{2}OA\cdot OB=\dfrac{2}{3}\Leftrightarrow\left|\dfrac{2-m}{m-1}\cdot\left(m-2\right)\right|=\dfrac{4}{3}\\ \Leftrightarrow\left|\dfrac{-\left(m-2\right)^2}{m-1}\right|=\dfrac{4}{3}\Leftrightarrow\left[{}\begin{matrix}\dfrac{-\left(m-2\right)^2}{m-1}=\dfrac{4}{3}\left(1\right)\\\dfrac{-\left(m-2\right)^2}{1-m}=\dfrac{4}{3}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow-3m^2+12m-12=4m-4\\ \Leftrightarrow3m^2-9m+9=0\\ \Leftrightarrow m\in\varnothing\\ \left(2\right)\Leftrightarrow-3m^2+12m-12=4-4m\\ \Leftrightarrow3m^2-16m+16=0\\ \Leftrightarrow\left[{}\begin{matrix}m=4\\m=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=4\\m=\dfrac{4}{3}\end{matrix}\right.\) thỏa đề
\(c,\) Gọi \(E\left(x_0;y_0\right)\) là điểm cần tìm
\(\Leftrightarrow\left(m-1\right)x_0+m-2=y_0\\ \Leftrightarrow mx_0+m-x_0-y_0-2=0\\ \Leftrightarrow m\left(x_o+1\right)-\left(x_0+y_0+2\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0=-1\\y_0=-2-x_0=-1\end{matrix}\right.\Leftrightarrow E\left(-1;-1\right)\)
a: Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\\left(m+1\right)x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x\left(m+1\right)=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{m+1}\end{matrix}\right.\)
vậy: \(A\left(-\dfrac{3}{m+1};0\right)\)
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=\left(m+1\right)\cdot x+3=0\left(m+1\right)+3=3\end{matrix}\right.\)
Vậy: B(0;3)
\(OA=\sqrt{\left(-\dfrac{3}{m+1}-0\right)^2+\left(0-0\right)^2}=\sqrt{\left(\dfrac{3}{m+1}\right)^2}=\left|\dfrac{3}{m+1}\right|\)
\(OB=\sqrt{\left(0-0\right)^2+\left(3-0\right)^2}=\sqrt{0+9}=3\)
Vì Ox\(\perp\)Oy
nên OA\(\perp\)OB
=>ΔOAB vuông tại O
=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot3\cdot\dfrac{3}{\left|m+1\right|}=\dfrac{9}{2\left|m+1\right|}\)
Để \(S_{AOB}=9\) thì \(\dfrac{9}{2\left|m+1\right|}=9\)
=>2|m+1|=1
=>|m+1|=1/2
=>\(\left[{}\begin{matrix}m+1=\dfrac{1}{2}\\m+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{1}{2}\\m=-\dfrac{3}{2}\end{matrix}\right.\)