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 \(x=y=\frac{1}{\sqrt{2}}\)

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.

\(\Leftrightarrow\left\{{}\begin{matrix}m-4< 0\\\Delta=\left(m+1\right)^2-4\left(m-4\right)\left(2m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\-7m^2+38m-15< 0\end{matrix}\right.\) \(\Rightarrow m< \frac{3}{7}\)

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)

2014

\(\Delta'=m^2-m^2+m-1=m-1\ge0\Rightarrow m\ge1\)

Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=-m+1\end{matrix}\right.\)

\(S=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=4m^2-2\left(-m+1\right)\)

\(=4m^2+2m+1\)

Xét \(f\left(m\right)=4m^2+2m+1\) trên \([1;+\infty)\)

\(a=4>0\) ; \(-\frac{b}{2a}=-\frac{1}{4}< 1\Rightarrow f\left(m\right)\) đồng biến trên \([1;+\infty)\)

\(\Rightarrow S_{min}=f\left(m\right)_{min}=f\left(1\right)=7\)

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: \(\left\{{}\begin{matrix}M+n=\frac{44}{7}\\Mn=\frac{20}{7}\end{matrix}\right.\)

\(\Rightarrow M^2+n^2=\left(M+n\right)^2-4Mn=\frac{1376}{49}\)

a,M=0
<=>(x-1)².(x+2)=0
=>TH1:x-1=0 <=> x=1
=>TH2:x+2=0<=> x=-2
Vậy với x=1 hoặc x=-2 thì M=0
b,M>0
<=>(x-1)².(x+2)>0
=>TH1: x-1 >0 ; x+2>0
<=> x>1 ; x>-2
=> x>1
=>TH2: x-1 <0 ; x+2<0
<=> x<1 ; x<-2
<=> x<-2
Vậy với x >1 hoặc x<-2 thì M>0
c,M<0
<=>(x-1)².(x+2)<0
=>TH1 : x-1 >0 ; x+2 <0
<=> x>1 ; x<-2
=> Không có giá trị x
=>TH2: x-1 <0 ; x+2 >0
<=> x<1 ; x>-2
=> -2<x<1
Vậy với -2<x<1 thì M<0