GL
1
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
CM
0
BA
0
NM
1
20 tháng 3 2020
Ta có: \(x \geqslant xy+1 \Rightarrow x-1 \geqslant xy\)
\( P = \dfrac{{3xy}}{{{x^2} + {y^2}}} = \dfrac{{3\left( {x - 1} \right)y + 3y}}{{{x^2} + {y^2}}}\\ \le \dfrac{{3x{y^2} + 3y}}{{2xy}} = \dfrac{{3y\left( {x + 3} \right)}}{{2xy}}\\ = \dfrac{{3\left( {x + 3} \right)}}{{2x}} = \dfrac{3}{2} + \dfrac{3}{{2x}} \le 2.\dfrac{3}{2} = 3\\ \Rightarrow {P_{\max }} = 3 \)
EC
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 3 2019
Nếu \(xy\le0\Rightarrow M\le0;\) nếu \(xy>0\Rightarrow M>0\Rightarrow\) GTLN nếu có của M sẽ xảy ra khi \(xy>0\)
Xét \(xy>0\Rightarrow xy+1>0\Rightarrow x>0\Rightarrow y>0\)
\(x\ge xy+1\Leftrightarrow1\ge y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\Rightarrow\frac{y}{x}\le\frac{1}{4}\) \(\Rightarrow\frac{x}{y}\ge4\)
\(M=\frac{3xy}{x^2+y^2}=\frac{3}{\frac{x}{y}+\frac{y}{x}}=\frac{3}{\frac{15}{16}.\frac{x}{y}+\frac{x}{16y}+\frac{y}{x}}\le\frac{3}{\frac{15}{16}.4+2\sqrt{\frac{xy}{16yx}}}=\frac{12}{17}\)
\(\Rightarrow M_{max}=\frac{12}{17}\) khi \(\left\{{}\begin{matrix}x=2\\y=\frac{1}{2}\end{matrix}\right.\)
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<=> \(\frac{\sqrt{x}}{2}=\frac{1}{2\sqrt{x}}\) => \(2\sqrt{x}=2\) => \(x=1\)
NT
1
HN
4 tháng 9 2018
\(P=\dfrac{xy}{x^2+y^2}=\dfrac{4}{17}-\dfrac{1}{17}\left(4x^2-17xy+4y^2\right)\le\dfrac{4}{17}-\dfrac{1}{17}\left[\left(\dfrac{x^2}{4}-2xy+4y^2\right)+\dfrac{15}{4}x^2-15x+15\right]\)
\(=\dfrac{4}{17}-\dfrac{1}{17}\left[\left(\dfrac{x}{2}-2y\right)^2+\dfrac{15}{4}\left(x-2\right)^2\right]\le\dfrac{4}{17}\)
Dâu = xảy ra khi \(x=2;y=\dfrac{1}{2}\)
HT
0
Max B =1/2 khi x =y = 1/2
có dung không?