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\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)
\(P_{max}=\dfrac{7}{2}\) khi \(x=-1\)
\(M=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}=2+\dfrac{1}{\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}}\le2+\dfrac{1}{\dfrac{3}{4}}=\dfrac{10}{3}\)
\(M_{max}=\dfrac{10}{3}\) khi \(x=-\dfrac{3}{2}\)
\(P_1=\frac{3x^2+6x+10}{x^2+2x+3}\)
\(=3+\frac{1}{x^2+2x+3}\)
Lại có: \(x^2+2x+3\)
\(=\left(x+1\right)^2+2\ge2\)
\(\Rightarrow P_1\le3+\frac{1}{2}=\frac{7}{2}\)
Dấu = xảy ra khi x=-1
P2 tương tự
\(M=-2x^2+2x-3\\ \Leftrightarrow2M=-4x^2+4x-6\\ \Leftrightarrow2M=-\left(4x^2-4x+4\right)-2\\ \Leftrightarrow2M=-\left(2x-2\right)^2-2\\ \Leftrightarrow M=\dfrac{-\left(2x-2\right)^2-2}{2}\)
Ta có :
\(-\left(2x-2\right)^2\le0\\ \Rightarrow-\left(2x-2\right)^2-2\le-2\\ \Rightarrow\dfrac{-\left(2x-2\right)^2-2}{2}\le\dfrac{-2}{2}\\ \Rightarrow M\le-1\)
\(\Rightarrow Max\left(M\right)=-1\Leftrightarrow2x-2=0\Rightarrow x=1\)
.......
\(N=3x-x^2-4\\ \Leftrightarrow N=-\left(x^2-3x+4\right)\\ \Leftrightarrow N=-\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}+\dfrac{16}{4}\right)\\ \Rightarrow N=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Ta có :
\(-\left(x-\dfrac{3}{2}\right)^2\le0\\ \Rightarrow-\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\le0+\dfrac{7}{4}\\ \Rightarrow N\le\dfrac{7}{4}\\ \Rightarrow Max\left(M\right)=\dfrac{7}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(P=\dfrac{3}{x^2-6x+10}=\dfrac{3}{\left(x-3\right)^2+1}\)
Ta có :
\(\left(x-3\right)^2\ge0\\ \Rightarrow\left(x-3\right)^2+1\ge1\\ \Rightarrow\dfrac{3}{\left(x-3\right)^2+1}\ge\dfrac{3}{1}\Rightarrow P\ge3\\ \Rightarrow Min\left(P\right)=3\Leftrightarrow x-3=0\Rightarrow x=3\)
a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
a/ \(M=x^2+y^2-x+6y+10=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+10-\frac{1}{4}-9\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Suy ra Min M = 3/4 <=> (x;y) = (1/2;-3)
b/
1/ \(A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Suy ra Min A = 7 <=> x = 2
2/ \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Suy ra Min B = 1/4 <=> x = 1/2
3/ \(N=2x-2x^2-5=-2\left(x^2-x+\frac{1}{4}\right)-5+\frac{1}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)
\(\ge-\frac{9}{2}\)
Suy ra Min N = -9/2 <=> x = 1/2
a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)
\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)
\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)
\(\Leftrightarrow9x-6=5x-20\)
\(\Leftrightarrow9x-5x=-20+6\)
\(\Leftrightarrow4x=-14\)
\(\Leftrightarrow x=-\dfrac{7}{2}\)
B3:\(\Rightarrow90.10^n-10^n.10^2+10^n.10-20\Rightarrow10^n.\left(90-10^2\right)+10^n.10-20\)
\(\Rightarrow10^n.\left(90-100\right)+10^n.10-20\Rightarrow-10.10^n+10^n.10-20\Rightarrow-20\)
\(A=-\left(x^2-x+5\right)=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{19}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\right]\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{19}{4}\le-\frac{19}{4}\)
Vậy \(A_{min}=-\frac{19}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
a,\(M=-2x^2+2x-3\)
\(\Rightarrow2M=-4x^2+4x-6=-\left(4x^2-4x+1\right)-5=-\left(2x-1\right)^2-5\)
Vì\(-\left(2x-1\right)^2\le0\Rightarrow2M=-\left(2x-1\right)^2-5\le-5\Rightarrow M\le-\frac{5}{2}\)
Dấu "=" xảy ra khi x=1/2
Vậy Mmax=-5/2 khi x=1/2
b, \(N=3x-x^2-4=-x^2+3x-4=-\left(x^2-3x+\frac{9}{4}\right)-\frac{7}{4}=-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\)
Vì \(-\left(x-\frac{3}{2}\right)^2\le0\Rightarrow N=-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)
Dấu "=" xảy ra khi x=3/2
Vậy Nmax=-7/4 khi x=3/2
c, \(P=\frac{3}{x^2-6x+10}=\frac{3}{x^2-6x+9+1}=\frac{3}{\left(x-3\right)^2+1}\)
Vì \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+1\ge1\Rightarrow\frac{1}{\left(x-3\right)^2+1}\le1\Rightarrow\frac{3}{\left(x-3\right)^2+1}\le3\)
Dấu "=" xảy ra khi x=3
Vậy Pmax=3 khi x=3