QV
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
23 tháng 12 2021
Câu 1:
\(ĐK:x\ge2\)
Áp dụng BĐT cauchy ta có:
\(\left(x+1\right)+4\ge2\sqrt{4\left(x+1\right)}=4\sqrt{x+1}\\ \Leftrightarrow2\sqrt{x+1}\le\dfrac{x+5}{2}\)
Ta có \(\left(x-2\right)+1\ge2\sqrt{x-2}\Leftrightarrow\sqrt{x-2}\le\dfrac{x-1}{2}\)
\(\Leftrightarrow P\le\dfrac{x+5}{2}+\dfrac{x-1}{2}-x+2013=x+2-x+2013=2015\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x-2=1\end{matrix}\right.\Leftrightarrow x=3\)
23 tháng 12 2021
Câu 2:
\(HPT\Leftrightarrow\left\{{}\begin{matrix}10\sqrt{x}+15y^3=140\\4y^3-10\sqrt{x}=12\end{matrix}\right.\left(x\ge0\right)\\ \Leftrightarrow19y^3=152\\ \Leftrightarrow y^3=8\Leftrightarrow y=2\\ \Leftrightarrow2\sqrt{x}+24=28\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Vậy \(\left(x;y\right)=\left(4;2\right)\)
Câu 3:
\(HPT\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\my+2m+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=\dfrac{3-2m}{m+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{m+1}\\x=\dfrac{3-2m}{m+1}\end{matrix}\right.\\ \Leftrightarrow xy=\dfrac{5\left(3-2m\right)}{\left(m+1\right)^2}\)
Đặt \(xy=t\)
\(\Leftrightarrow m^2t+2mt+t=15-10m\\ \Leftrightarrow m^2t+2m\left(t+5\right)+t-15=0\)
PT có nghiệm nên \(\Delta'=\left(t+5\right)^2-t\left(t-15\right)\ge0\)
\(\Leftrightarrow10t+25+15t\ge0\Leftrightarrow t\ge-1\)
Vậy \(xy_{min}=-1\Leftrightarrow\dfrac{5\left(2m-3\right)}{\left(m+1\right)^2}=1\Leftrightarrow m^2-8m+16=0\Leftrightarrow m=4\)
22 tháng 11 2021
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
TN
16 tháng 11 2016
Bài 1:
\(P=x\sqrt{3-x^2}=\sqrt{x^2}\cdot\sqrt{3-x^2}\)
\(=\sqrt{x^2\left(3-x^2\right)}\)\(\le\frac{x^2+3-x^2}{2}=\frac{3}{2}\)
Dấu = khi \(x=\sqrt{\frac{3}{2}}\)
Vậy MaxP=\(\frac{3}{2}\Leftrightarrow x=\sqrt{\frac{3}{2}}\)
K
2 tháng 8 2020
Bài 2 :
Tìm min : Bình phương
Tìm max : Dùng B.C.S ( bunhiacopxki )
Bài 3 : Dùng B.C.S
2 tháng 8 2020
KP9
nói thế thì đừng làm cho nhanh bạn ạ
Người ta cũng có chút tôn trọng lẫn nhau nhé đừng có vì dăm ba cái tích
28 tháng 8 2019
\(A=\sqrt{x^2+y\left(y-2x\right)}+\sqrt{y^2+z\left(z-2y\right)}+\sqrt{x^2+z\left(z-2x\right)}\)
\(=\sqrt{x^2-2xy+y^2}+\sqrt{y^2-2yz-z^2}+\sqrt{x^2-2xz+z^2}\)
\(=\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-z\right)^2}+\sqrt{\left(z-x\right)^2}\)
\(=x-y+y-z+z-x\)
\(=0\)
7 tháng 7 2021
\(a,B=\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}\right):\left(\frac{1-xy+x+y+2xy}{1-xy}\right)\)
\(B=\frac{\sqrt{x}+\sqrt{y}+x\sqrt{y}+y\sqrt{x}+\sqrt{x}-\sqrt{y}-x\sqrt{y}+y\sqrt{x}}{1-xy}.\frac{1-xy}{1+xy+x+y}\)
\(B=\frac{2\sqrt{x}+2y\sqrt{x}}{x\left(y+1\right)+\left(y+1\right)}\)
\(B=\frac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}\)
\(B=\frac{2\sqrt{x}}{x+1}\)
\(b,B=\frac{2\sqrt{\frac{2}{2+\sqrt{3}}}}{\frac{2}{2+\sqrt{3}}+1}\)
\(\frac{2\sqrt{\frac{4}{4+2\sqrt{3}}}}{\frac{4}{4+2\sqrt{3}}+1}\)
\(B=\frac{2\sqrt{\frac{4}{\left(\sqrt{3}+1\right)^2}}}{\frac{4}{\left(\sqrt{3}+1\right)^2}+1}\)
\(B=\frac{2.2}{\sqrt{3}+1}:\frac{4+2\sqrt{3}}{\sqrt{3}+1}\)
\(B=\frac{4}{\left(\sqrt{3}+1\right)^2}\)
\(B=\left(\frac{2}{\sqrt{3}+1}\right)^2\)
\(c,B=\frac{2\sqrt{x}}{x+1}\)
\(B=\frac{2}{\sqrt{x}+\frac{1}{\sqrt{x}}}\)
ta có :
\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)
dấu "=" xảy ra khi \(x=1\)
\(< =>MAX:B=\frac{2}{2}=1\)
7 tháng 7 2021
Đk: x \(\ge\)0; y \(\ge\)0; xy \(\ne\)1
Ta có: B = \(\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\frac{x+y+2xy}{1-xy}\right)\)
B = \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\frac{1-xy+x+y+2xy}{1-xy}\)
B = \(\frac{x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}\cdot\frac{1-xy}{x+y+xy+1}\)
B = \(\frac{2\sqrt{x}+2y\sqrt{x}}{\left(y+1\right)\left(x+1\right)}=\frac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\frac{2\sqrt{x}}{x+1}\)
b) Ta có: \(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4-2\sqrt{3}}{4-3}=4-2\sqrt{3}\)
=> \(x=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)=> \(\sqrt{x}=\sqrt{3}-1\)
Do đó, B = \(\frac{2.\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\frac{2\sqrt{3}-2}{5-2\sqrt{3}}=\frac{\left(2\sqrt{3}-2\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}=\frac{10\sqrt{3}+12-10-4\sqrt{3}}{25-12}\)
B = \(\frac{6\sqrt{3}+2}{13}\)
c) Ta có: \(\frac{1}{B}=\frac{x+1}{2\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\ge2\cdot\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{2\sqrt{x}}}=2\cdot\sqrt{\frac{1}{4}}=1\)(đk: x \(\ne\)0)
=> \(B\le\frac{1}{1}=1\)Dấu "==" xảy ra<=> \(\frac{\sqrt{x}}{2}=\frac{1}{2\sqrt{x}}\) => \(2\sqrt{x}=2\) => \(x=1\)