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BĐT Bu nhi a cốp xki :
\(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(\Rightarrow\left(x.1+y.1\right)^2\le\left(1^2+1^2\right)\left(x^2+y^2\right)\)
\(\Rightarrow\left(x+y\right)^2\le2\left(x^2+y^2\right)\)
\(\Rightarrow x+y\le\sqrt{2\left(x^2+y^2\right)}\)Nguyễn Thị Thanh Trang
\(P=2018xy+2019\left(x+y\right)\le2018.\frac{x^2+y^2}{2}+2019\sqrt{2\left(x^2+y^2\right)}=2018.\frac{1}{2}+2019\sqrt{2.1}=1009+2019\sqrt{2}\)
Vậy GTLN của P là \(1009+2019\sqrt{2}\) . Dấu \("="\) xảy ra khi \(x=y=\frac{1}{\sqrt{2}}\)
\(P=\frac{2}{3xy}+\frac{3}{\sqrt{3\left(1+y\right)}}\ge\frac{2}{3y\left(3-y\right)}+\frac{6}{y+4}\)
\(\Rightarrow P\ge2\left(\frac{-9y^2+28y+4}{3\left(-y^3-y^2+12y\right)}\right)=2\left(\frac{2\left(-y^3-y^2+12y\right)+2y^3-7y^2+4y+4}{3\left(-y^3-y^2+12y\right)}\right)\)
\(P\ge2\left(\frac{2}{3}+\frac{\left(y-2\right)^2\left(2y+1\right)}{3y\left(3-y\right)\left(y+4\right)}\right)\ge\frac{4}{3}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
@Nguyễn Việt Lâm duyệt bài giúp em với ạ @Phạm Minh Quang nick đây
Ta có \(-\frac{b}{2a}=\frac{3}{2}\in\left[-2;3\right]\)
\(y\left(-2\right)=-5\) ; \(y\left(\frac{3}{2}\right)=-54\); \(y\left(3\right)=-45\)
\(\Rightarrow M=-5\) ; \(m=-54\)
a)x2+6x+10
=x2+2.3x+32+1
=(x+3)2+1
Vì (x+3)2\(\ge\)0
Suy ra:(x+3)2+1\(\ge\)1(đpcm)
b)9x2-6x+2
=(3x)2-2.3x+12+1
=(3x-1)2+1
Vì (3x-1)2\(\ge\)0
Suy ra:(3x-1)2+1\(\ge\)1(đpcm)
c)x2+x+1
=x2+2.\(\frac{1}{2}x\)+\(\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
=\(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\)
Suy ra:\(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\left(đpcm\right)\)
d)3x2+3x+1
Ta có:Vì 3x2 là số nguyên dương
Mà x2>x
Suy ra:3x2-3x là số nguyên dương
Vậy 3x2+3x+1 là số nguyên dương(đpcm)
a.
\(\Leftrightarrow\Delta'=4\left(m+1\right)^2+1-m^2< 0\)
\(\Leftrightarrow3m^2+8m+5< 0\Rightarrow-\frac{5}{3}< x< -1\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}m-4< 0\\\Delta=\left(m+1\right)^2-4\left(m-4\right)\left(2m-1\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\-7m^2+38m-15< 0\end{matrix}\right.\) \(\Rightarrow m< \frac{3}{7}\)
Đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)
\(P=\frac{2-2sina.cosa+cos^2a}{4sin^2a-3sina.cosa+cos^2a}=\frac{2-sin2a+\frac{1+cos2a}{2}}{1+\frac{3\left(1-cos2a\right)}{2}-\frac{3}{2}sin2a}=\frac{5-2sin2a+cos2a}{5-3cos2a-3sin2a}\)
\(\Leftrightarrow3P-3P.cos2a-3P.sin2a=5-2sin2a+cos2a\)
\(\Leftrightarrow\left(3P-2\right)sin2a+\left(3P+1\right)cos2a=5P-5\)
Áp dụng BĐT Bunhiacopxki:
\(\left(5P-5\right)^2\le\left(3P-2\right)^2+\left(3P+1\right)^2\)
\(\Leftrightarrow7P^2-44P+20\le0\)
Theo Viet: \(\left\{{}\begin{matrix}M+n=\frac{44}{7}\\Mn=\frac{20}{7}\end{matrix}\right.\)
\(\Rightarrow M^2+n^2=\left(M+n\right)^2-4Mn=\frac{1376}{49}\)