Cho x,y là các số thực
CMR: \(-\frac{1}{4}\le\frac{\left(x^2-y^2\right)\left(1x^2y^2\right)}{\left(1+x^2\right)\left(1+y^2\right)^2}\le\frac{1}{4}\)
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Cho x,y là các số thực. CMR
\(-\frac{1}{4}\le\frac{\left(x^2-y^2\right)\left(1-x^2y^2\right)}{\left(1+x^2\right)^2\left(1+y^2\right)^2}\le\frac{1}{4}\)
Cho x,y là các số thực. CMR \(-\frac{1}{4}\le\frac{\left(x^2-y^2\right)\left(1-x^2y^2\right)}{\left(1+x^2\right)^2\:\left(1+y^2\right)^2}\le\frac{1}{4}\)
M^2left(sqrt{x}+sqrt{2y}right)^2left(frac{1}{_{sqrt{alpha}}}.sqrt{alpha x}+sqrt{2y}right)^2 left(frac{1}{alpha}+1right)left(alpha x+2yright)Rightarrow M^4leleft(frac{1}{alpha}+1right)^2left(alpha x+2yright)^2leleft(frac{1}{alpha}+1right)^2left(alpha^2+4right)left(x^2+y^2right)left(frac{1}{alpha}+1right)^2left(alpha^2+4right)Dấu bằng xảy ra hept{begin{cases}frac{alpha x}{frac{1}{alpha}}frac{2y}{1}frac{alpha}{x}frac{2}{y}end{cases}}Rightarrowhept{begin{cases}alpha^2x2yalphafrac{2x}{y}end{cases}R...
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\(M^2=\left(\sqrt{x}+\sqrt{2y}\right)^2=\left(\frac{1}{_{\sqrt{\alpha}}}.\sqrt{\alpha x}+\sqrt{2y}\right)^2< =\left(\frac{1}{\alpha}+1\right)\left(\alpha x+2y\right)\)
\(\Rightarrow M^4\le\left(\frac{1}{\alpha}+1\right)^2\left(\alpha x+2y\right)^2\le\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)\left(x^2+y^2\right)=\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)\)
Dấu bằng xảy ra => \(\hept{\begin{cases}\frac{\alpha x}{\frac{1}{\alpha}}=\frac{2y}{1}\\\frac{\alpha}{x}=\frac{2}{y}\end{cases}}\Rightarrow\hept{\begin{cases}\alpha^2x=2y\\\alpha=\frac{2x}{y}\end{cases}\Rightarrow\hept{\begin{cases}\frac{\alpha^2}{2}=\frac{y}{x}\\\frac{\alpha}{2}=\frac{x}{y}\end{cases}}}\Rightarrow\frac{\alpha^2}{2}=\frac{1}{\frac{\alpha}{2}}\Rightarrow\alpha=\sqrt[3]{4}\)
Suy ra max = \(\sqrt[4]{\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)}\) với \(\alpha=\sqrt[3]{4}\)
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Cho các số thực x, y, z thõa mãn xyz = 1. Chứng minh rằng:
\(\frac{1}{\left(2+x\right)\left(2+\frac{1}{y}\right)}+\frac{1}{\left(2+y\right)\left(2+\frac{1}{z}\right)}+\frac{1}{\left(2+z\right)\left(2+\frac{1}{x}\right)}\le\frac{1}{3}\)
\(\Sigma\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\Sigma\left(\dfrac{1}{9}.\dfrac{a^2\left(2+1\right)^2}{2a.\left(\Sigma a\right)+2a^2+bc}\right)\le\Sigma\left(\dfrac{1}{9}.\dfrac{4a^2}{2a\left(\Sigma a\right)}+\dfrac{1}{9}.\dfrac{a^2}{2a^2+bc}\right)\)
\(=\Sigma\left(\dfrac{1}{9}.\left(\dfrac{2a}{\Sigma a}+\dfrac{a^2}{2a^2+bc}\right)\right)=\dfrac{1}{9}\left(2+\Sigma\dfrac{a^2}{2a^2+bc}\right)\)
Cần chứng minh \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
<=> \(\Sigma\frac{bc}{2a^2+bc}\ge1\) (*)
Đặt (x;y;z) -------> \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)
Suy ra (*) <=> \(\Sigma\frac{x^2}{x^2+2xy}\ge1\Leftrightarrow\frac{\Sigma x^2}{\Sigma x^2}\ge1\) (đúng)
Vậy \(\Sigma\frac{a^2}{2a^2+bc}\le1\)
Suy ra \(\Sigma\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\le\frac{1}{9}\left(2+\Sigma\frac{a^2}{2a^2+bc}\right)\le\frac{1}{9}\left(2+1\right)=\frac{1}{3}\)
Đẳng thức xảy ra <=> x = y = z = 1
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Cho x,y,z là 3 số thực dương thỏa mãn xyz=1.Chứng minh bất đẳng thức
\(\frac{1}{\left(2x+y+z\right)^2}+\frac{1}{\left(x+2y+z\right)^2}+\frac{1}{\left(x+y+2z\right)^2}\le\frac{3}{16}\)
Cho \(0\le x\le2;0\le y\le\frac{1}{2}\).Chứng minh rằng \(\left(2x-x^2\right)\left(y-2y^2\right)\le\frac{1}{8}\)
giải hệ phương trình
1 , left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)+2xyleft(y-xright)left(y-1right)left(y+xright)left(y-2right)-2xyend{matrix}right.
2, left{{}begin{matrix}2left(frac{1}{x}+frac{1}{2y}right)+3left(frac{1}{x}-frac{1}{2y}right)^29left(frac{1}{x}+frac{1}{2y}right)-6left(frac{1}{x}-frac{1}{2y}right)^2-3end{matrix}right.
3 , left{{}begin{matrix}frac{xy}{x+y}frac{2}{3}frac{yz}{y+z}frac{6}{5}frac{zx}{z+x}frac{3}{4}end{matrix}right.
4 , left{{}begin...
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giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
Áp dụng BĐT Cô-si để tìm Maxa. yleft(x+3right)left(5-xright),left(-3le xle5right) b. yxleft(6-xright)left(0le xle6right) c. yleft(x+3right)left(5-2xright)left(-3le xlefrac{5}{2}right) d. yleft(2x+5right)left(5-2xright)left(-frac{5}{2}le xle5right) e. yleft(6x+3right)left(5-2xright)left(-frac{1}{2}le xlefrac{5}{2}right) f. yfrac{x}{x^2+2},xge0 g. yfrac{x^2}{left(x^2+2right)^3}
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Áp dụng BĐT Cô-si để tìm Max
a. \(y=\left(x+3\right)\left(5-x\right),\left(-3\le x\le5\right)\)
b. \(y=x\left(6-x\right)\left(0\le x\le6\right)\)
c. \(y=\left(x+3\right)\left(5-2x\right)\left(-3\le x\le\frac{5}{2}\right)\)
d. \(y=\left(2x+5\right)\left(5-2x\right)\left(-\frac{5}{2}\le x\le5\right)\)
e. \(y=\left(6x+3\right)\left(5-2x\right)\left(-\frac{1}{2}\le x\le\frac{5}{2}\right)\)
f. \(y=\frac{x}{x^2+2},x\ge0\)
g. \(y=\frac{x^2}{\left(x^2+2\right)^3}\)
Từ bđt Cauchy : \(a+b\ge2\sqrt{ab}\) ta suy ra được \(ab\le\frac{\left(a+b\right)^2}{4}\)
Áp dụng vào bài toán của bạn :
a/ \(y=\left(x+3\right)\left(5-x\right)\le\frac{\left(x+3+5-x\right)^2}{4}=...............\)
b/ Tương tự
c/ \(y=\left(x+3\right)\left(5-2x\right)=\frac{1}{2}.\left(2x+6\right)\left(5-2x\right)\le\frac{1}{2}.\frac{\left(2x+6+5-2x\right)^2}{4}=.............\)
d/ Tương tự
e/ \(y=\left(6x+3\right)\left(5-2x\right)=3\left(2x+1\right)\left(5-2x\right)\le3.\frac{\left(2x+1+5-2x\right)^2}{4}=.......\)
f/ Xét \(\frac{1}{y}=\frac{x^2+2}{x}=x+\frac{2}{x}\ge2\sqrt{x.\frac{2}{x}}=2\sqrt{2}\)
Suy ra \(y\le\frac{1}{2\sqrt{2}}\)
..........................
g/ Đặt \(t=x^2\) , \(t>0\) (Vì nếu t = 0 thì y = 0)
\(\frac{1}{y}=\frac{t^3+6t^2+12t+8}{t}=t^2+6t+\frac{8}{t}+12\)
\(=t^2+6t+\frac{8}{3t}+\frac{8}{3t}+\frac{8}{3t}+12\)
\(\ge5.\sqrt[5]{t^2.6t.\left(\frac{8}{3t}\right)^3}+12=.................\)
Từ đó đảo ngược y lại rồi đổi dấu \(\ge\) thành \(\le\)
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cm các biểu thức sau ko phụ thuộc vào biến:
a,\(\left[\frac{2\left(x+1\right)\left(y+1\right)}{\left(x+1\right)^2-\left(y+1\right)^2}+\frac{x-y}{2x+2y+4}\right].\frac{2x+2}{x+y+2}+\frac{y+1}{y-x}\)
b,\(\left[2\left(x+y\right)+1-\frac{1}{1-2x-2y}\right]:\left[2x+2y-\frac{4x^2+8xy+4y^2}{2x+2y-1}\right]+2\left(x+y\right)\)