Cho \(-1\le x\le1\). Tìm GTNN của biểu thức \(y=\frac{5-3x}{\sqrt{1-x^2}}\)
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Khách
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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 \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
LH
0
TN
22 tháng 1 2017
Ta có: \(1\ge x+y\ge2\sqrt{xy}\Rightarrow1\ge4xy\Rightarrow\frac{1}{xy}\ge4\)
\(\Rightarrow P\ge2\sqrt{\frac{1}{xy}}\cdot\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}\)
Mà \(\frac{1}{xy}+xy=\frac{15}{16}\cdot\frac{1}{xy}+\frac{1}{16xy}+xy\)
\(\ge\frac{15}{16}\cdot4+2\sqrt{\frac{1}{16xy}\cdot xy}=\frac{15}{16}\cdot4+\frac{2}{4}=\frac{17}{4}\)
\(\Rightarrow P\ge2\cdot\frac{\sqrt{17}}{2}=\sqrt{17}\) xảy ra khi \(x=y=\frac{1}{2}\)
21 tháng 7 2020
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
21 tháng 7 2020
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 5 2019
\(2\sqrt{xy}\le x+y\le1\Rightarrow\frac{1}{\sqrt{xy}}\ge2\Rightarrow\frac{1}{xy}\ge4\)
\(P\ge\frac{2}{\sqrt{xy}}\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}=2\sqrt{\frac{15}{16xy}+\frac{1}{16xy}+xy}\)
\(P\ge2\sqrt{\frac{15}{16}.4+2\sqrt{\frac{xy}{16xy}}}=\sqrt{17}\)
\(\Rightarrow P_{min}=\sqrt{17}\) khi \(x=y=\frac{1}{2}\)
22 tháng 2 2017
\(A^2=\left(2\sqrt{x-4}+\sqrt{8-x}\right)^2\le\left(2^2+1^2\right)\left(x-4+8-x\right)=20..\)
\(A\le2\sqrt{5}..\)
9 tháng 11 2016
a/ \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}.\frac{18}{x}}=...\)
b/ \(\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}.\frac{2}{x-1}}+\frac{1}{2}=...\)
c/ \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=...\)
d/ \(\frac{x}{3}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=...\)
e/ \(\frac{x}{1-x}+\frac{5}{x}=\frac{x}{1-x}+\frac{5-5x+5x}{x}=\frac{x}{1-x}+\frac{5\left(1-x\right)}{x}+5\ge2\sqrt{\frac{x}{1-x}.\frac{5\left(1-x\right)}{x}}+5=...\)
f/ \(\frac{x^3+1}{x^2}=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge2\sqrt{\frac{x}{2}.\frac{x}{2}.\frac{1}{x^2}}=...\)
g/ \(\frac{x^2+4x+4}{x}=x+\frac{4}{x}+4\ge2\sqrt{x.\frac{4}{x}}+4=...\)
Đặt \(a=\sqrt{1-x},a\ge0\) ; \(b=\sqrt{1+x},b\ge0\)
\(\Rightarrow y=\frac{5-3x}{\sqrt{1-x^2}}=\frac{\left(1+x\right)+4\left(1-x\right)}{\sqrt{1+x}.\sqrt{1-x}}=\frac{b^2+4a^2}{ab}\)
Áp dụng bất đẳng thức Cauchy , ta có : \(\frac{b^2+4a^2}{ab}\ge\frac{2.\sqrt{b^2.4a^2}}{ab}=\frac{4ab}{ab}=4\)
Dấu đẳng thức xảy ra \(\Leftrightarrow b^2=4a^2\Leftrightarrow b=2a\Leftrightarrow\sqrt{1+x}=2\sqrt{1-x}\Leftrightarrow x=\frac{3}{5}\)
Vậy Min y = 4 \(\Leftrightarrow x=\frac{3}{5}\)