Min P= (x^2 -2x + 2016)/ (x^2)
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x^2-2x+2016=(x-1)^2+2015>=2015
=> min của x^2-2x+2016=2015 khi x =1
-x^2+2x+2016=-(x-1)^2+2017=<2017
=> max -x^2+2x+2016 =2017 khi x=1
min của B = 2016
= 0^2-2x0+2016
= 0-0+2016
khi x = 0 (vì min: nhỏ nhất)
ủng hộ nhé
x khac 0
Bx^2=x^2-2x+2016
(1-B)x^2-2x+2016=0
\(\Rightarrow\Delta=1-4.\left(1-B\right).2016\ge0\Rightarrow1-4.2016+4.2016B\ge0\)
\(B\ge\frac{4.2016-1}{4.2016}=1-\frac{1}{4.2016}\)
GTNN(B)=1-1/(4.2016)
bắt hết các loại gió mùa
Ta có:
\(B=\frac{x^2-2x+2016}{x^2}\Rightarrow2016B=\frac{2015x^2+\left(x^2-2.2016x+2016^2\right)}{x^2}=2015+\frac{\left(x-2016\right)^2}{x^2}\ge2015\)
Dấu "=" xảy ra khi \(\frac{\left(x-2016\right)^2}{x^2}=0\Rightarrow x=2016\)
\(\Rightarrow2016B_{min}=2015\Rightarrow B_{min}=\frac{2015}{2016}\) khi \(x=2016\)
\(B=\dfrac{x^2-2x+2016}{x^2}\\ \\ =\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{2016}{x^2}\\ \\ =1-\dfrac{2}{x}+\dfrac{2016}{x^2}\\ =\dfrac{2016}{x^2}-\dfrac{2}{x}+\dfrac{1}{2016}+\dfrac{2015}{2016}\\ =\left(\dfrac{2016}{x^2}-\dfrac{2}{x}+\dfrac{1}{2016}\right)+\dfrac{2015}{2016}\\ =2016\left(\dfrac{1}{x^2}-\dfrac{1}{1008x}+\dfrac{1}{2016^2}\right)+\dfrac{2015}{2016}\\ =2016\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2+\dfrac{2015}{2016}\)
Do \(2016\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2\ge0\forall x\)
\(\Rightarrow B=2016\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2+\dfrac{2015}{2016}\ge\dfrac{2015}{2016}\forall x\)
Dấu "=" xảy ra khi:
\(2016\left(\dfrac{1}{x}-\dfrac{1}{2016}\right)^2=0\\ \Leftrightarrow\dfrac{1}{x}-\dfrac{1}{2016}=0\\ \Leftrightarrow\dfrac{1}{x}=\dfrac{1}{2016}\\ \Leftrightarrow x=2016\)
Vậy \(B_{Min}=\dfrac{2015}{2016}\) khi \(x=2016\)
a)\(M=x^2-2xy+2y^2-4y+2016\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012\)
\(=\left(x-y\right)^2+\left(y-2\right)^2+2012\ge2012\)
Dấu = khi \(\begin{cases}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y=0\\y-2=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=y\\y=2\end{cases}\)\(\Leftrightarrow x=y=2\)
Vậy MinM=2012 khi x=y=2
b)\(N=x^2-2xy+2x+2y^2-4y+2016\)
\(=\left(x^2-2xy+2x+y^2-2y+1\right)+\left(y^2-2y+1\right)+2014\)
\(=\left(x-y+1\right)^2+\left(y-1\right)^2+2014\ge2014\)
Dấu = khi \(\begin{cases}\left(x-y+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y+1=0\\y-1=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x-y+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x-1+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x=0\\y=1\end{cases}\)
Vậy MinN=2014 khi x=0;y=1
A=x2+2x+2016=(x2+2x+1)+2015=(x+1)2+2015
ta thấy : (x+1)2>=0
=>A>=2015
=> GTNN của A=2015 khi x=-1
B=-x2+2x+2016=-(x2-2x+1)+2017=2017-(x-1)2
ta thấy :-(x-1)2<=0
=> GTLN của B=2017 khi x=1
Có :
P = 1 - 2/x + 2016/x^2
= (2016/x^2 - 2/x + 1/2016) + 2015/2016
= \(\left(\frac{\sqrt{2016}}{x}-\frac{1}{\sqrt{2016}}\right)\)^ 2 + 2015/2016 >= 2015/2016
Dấu "=" xảy ra <=> \(\frac{\sqrt{2016}}{x}-\frac{1}{\sqrt{2016}}=0\)<=> x=2016
Vậy ..............
Tk mk nha