Tìm Max Q=\(\dfrac{x}{\left(2015\right)^2}\);x>0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
T
22 tháng 7 2016
a Tách \(M=2+\frac{4xy}{x^2+2xy+y^2}=2+\frac{4xy}{\left(x+y\right)^2}\le2+1=3\)
Dấu = xảy ra khi và chỉ khi x=y và x+y=2015 <=>x=y=2015/2
b,:\(N\ge\frac{\left(1+\frac{2015}{x}+1+\frac{2015}{y}\right)^2}{2}=\frac{\left(2+2015\left(\frac{1}{x}+\frac{1}{y}\right)\right)^2}{2}\)
áp dunngj svac =>\(N\ge\frac{\left(2+2015\left(\frac{\left(1+1\right)^2}{x+y}\right)\right)^2}{2}=\frac{\left(2+\frac{2015.4}{2015}\right)^2}{2}=18\)
dấu = xảy ra khi và chỉ khi x=y và x+y=2015 <=>x=y=2015/2
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 12 2018
\(VT=\dfrac{2}{x}-\dfrac{2}{x+1}+\dfrac{2}{x+1}-\dfrac{2}{x+2}+...+\dfrac{2}{x+2014}-\dfrac{2}{x+2015}\)
\(VT=\dfrac{2}{x}-\dfrac{2}{x+2015}=\dfrac{2\left(x+2015-x\right)}{x\left(x+2015\right)}=\dfrac{4030}{x\left(x+2015\right)}\)
NT
1
24 tháng 5 2015
\(A=\frac{x+2015-2015}{\left(x+2015\right)^2}=\frac{1}{x+2015}-\frac{2015}{\left(x+2015\right)^2}\)
Đặt \(y=\frac{1}{x+2015}\)=> \(A=y-2015y^2\)
\(A=-2015\left(y^2-2.\frac{1}{4030}.y+\frac{1}{4030^2}\right)+\frac{2015}{4030^2}=-2015\left(y-\frac{1}{4030}\right)^2+\frac{1}{8060}\le0+\frac{1}{8060}\)
=> A max = 1/8060 khi \(y=\frac{1}{4030}\Rightarrow\frac{1}{x+2015}=\frac{1}{4030}\Rightarrow x+2015=4030\)=> x = 2015
DN
24 tháng 5 2018
Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)
9 tháng 8 2017
sai dề kìa \(\frac{6x+3}{x^3+1}\)mới đúng
ĐK : \(x\ne-1\)
a) rút gọn được \(C=\frac{1}{x^2-x+1}\)
b)\(C=\frac{1}{3}\Rightarrow\frac{1}{x^2-x+1}=\frac{1}{3}\)
\(\Rightarrow x^2-x+1=3\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)=0\\\left(x-2\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\left(Loai\right)\\x=2\left(Nhan\right)\end{cases}}}\)
vậy khi \(C=\frac{1}{3}\)thì x=2
c)\(C=\frac{1}{x^2-x+2}\)
ta có \(x^2-x+2=x^2-2x\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
\(\Rightarrow C=\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{7}{4}\)
vậy max \(C=\frac{7}{4}\)khi và chỉ khi \(x=\frac{1}{2}\)
DT
0
24 tháng 7 2017
\(E=\dfrac{4\left|x\right|+9}{\left|x\right|+1}\)
\(\left\{{}\begin{matrix} \left|x\right|\ge0\Rightarrow4\left|x\right|\ge0\Rightarrow4\left|x\right|+9\ge9\\\left|x\right|\ge0\Rightarrow x+1\ge1\end{matrix}\right.\)
\(MAX_E\Rightarrow MIN_{\left|x\right|+1}\)
\(MIN_{\left|x\right|+1}=1\)
\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)
\(\Rightarrow MAX_E=\dfrac{4.\left|0\right|+9}{\left|0\right|+1}=\dfrac{9}{1}=9\)
\(F=\dfrac{2\left|x\right|+8}{3\left|x\right|+1}\)
\(\left\{{}\begin{matrix}\left|x\right|\ge0\Rightarrow2\left|x\right|\ge0\Rightarrow2\left|x\right|+8\ge8\\\left|x\right|\ge0\Rightarrow3\left|x\right|\ge0\Rightarrow3\left|x\right|+1\ge1\end{matrix}\right.\)
\(MAX_F\Rightarrow MIN_{3\left|x\right|+1}\)
\(MIN_{3\left|x\right|+1}=1\)
\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)
\(\Rightarrow MAX_F=\dfrac{2.\left|0\right|+8}{3.\left|0\right|+1}=\dfrac{8}{1}=8\)
\(\)
AH
Akai Haruma
Giáo viên
19 tháng 4 2018
Lời giải:
Ta có:
\(f(x)=x^2+x\Rightarrow \frac{1}{f(x)}=\frac{1}{x^2+x}=\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\)
Do đó:
\(\frac{1}{f(1)}=1-\frac{1}{2}\)
\(\frac{1}{f(2)}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{f(3)}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{f(2014)}=\frac{1}{2014}-\frac{1}{2015}\)
\(\frac{1}{f(2015)}=\frac{1}{2015}-\frac{1}{2016}\)
Cộng theo vế:
\(\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(2014)}+\frac{1}{f(2015)}=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
?