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\(A\left(x\right)=-\left(x^2-\frac{5}{3}x\right)+1=-3\left(x^2-2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2\right)+1+3.\left(\frac{5}{6}\right)^2\)
\(=-3\left(x-\frac{5}{6}\right)^2+\frac{37}{12}\le\frac{37}{12}\)
Dấu "=" xảy ra khi \(x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)
Vậy GTLN của A là 37/12.
b, c làm tương tự.
\(1,A=x\left(x+1\right)+5\)
\(=x^2+x+5\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\)
Ta có : \(\left(x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dâu = xảy ra \(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(Min_A=\dfrac{19}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
\(2,B=-x^2-4x+9\)
\(=-\left(x^2+4x+4\right)+13\)
\(=-\left(x+2\right)^2+13\)
Ta có :\(\left(x+2\right)^2\ge0\Rightarrow-\left(x+2\right)^2\le0\Rightarrow-\left(x+2\right)^2+13\le13\)
Dấu = xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy \(Max_B=13\Leftrightarrow x=-2\)
\(3,C=x^2-4x+7+y^2+2y\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+2\)
\(=\left(x-2\right)^2+\left(y+1\right)^2+2\)
Ta có :
\(\left(x-2\right)^2\ge0;\left(y+1\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+2\ge2\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy \(Min_C=2\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
a) \(x\left(x+1\right)+5\)
\(=x^2+x+5\)
\(=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}\)
\(=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{19}{4}\)
\(=\left[x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{19}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\)
Vậy GTNN của biểu thức trên bằng \(\dfrac{19}{4}\) khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{-1}{2}\)
b) \(-x^2-4x+9\)
\(=-x^2-4x-4+13\)
\(=-\left(x^2+4x+4\right)+13\)
\(=-\left(x^2+2.x.2+2^2\right)+13\)
\(=-\left(x+2\right)^2+13\)
Vậy GTLN của biểu thức trên bằng \(13\) khi \(x+2=0\Leftrightarrow x=-2\)
\(a)\)
\(A=2x^2+x\)
\(\Leftrightarrow A=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
\(MinA=\frac{-1}{8}\)khi \(x=\frac{-1}{4}\)
\(b)\)
\(B=x^2+2x+y^2-4y+6\)
\(\Leftrightarrow B=x^2+2x+1+y^2-4y+4+1\)
\(\Leftrightarrow B=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\)
Dấu '' = '' xảy ra khi: \(x=-1;y=2\)
\(c)\)
\(C=4x^2+4x+9y^2-6y-5\)
\(\Leftrightarrow C=4x^2+4x+1+9y^2-6y+1-7\)
\(\Leftrightarrow C=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
Dấu '' = '' xáy ra khi: \(x=\frac{-1}{2};y=\frac{1}{3}\)
\(C=2\left(x-\frac{5}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\Rightarrow C_{min}=\frac{7}{8}\)
\(D=\left(x^2+4xy+4y^2\right)+\left(y^2+y+\frac{1}{4}\right)+\frac{8083}{4}\)
\(D=\left(x+2y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{8083}{4}\ge\frac{8083}{4}\)
\(E=\frac{1}{2}\left(4x^2+y^2+\frac{9}{4}-4xy-6x+3y\right)+\frac{1}{2}\left(y^2+y+\frac{1}{4}\right)+\frac{15}{4}\)
\(E=\frac{1}{2}\left(2x-y-\frac{3}{2}\right)^2+\frac{1}{2}\left(y+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\)
\(A=-\left(x-2\right)^2+11\le11\)
\(B=-\left(x+\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)
\(C=-\left(x-3y\right)^2-\left(y-2\right)^2+11\le11\)
\(a,M=x^2-4x+5=\left(x-2\right)^2+5\\ \Rightarrow M\ge5\)
Dấu "=" xảy ra \(\Leftrightarrow x=2\)
\(b,N=y^2-y-3=\left(y-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\\ \Rightarrow N\ge-\dfrac{13}{4} \)
Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)
\(P=x^2+y^2-4x+y+7=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ \Rightarrow P\ge\dfrac{11}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)
a: M=x^2-4x+4+1
=(x-2)^2+1>=1
Dấu = xảy ra khi x=2
b: N=y^2-y+1/4-13/4
=(y-1/2)^2-13/4>=-13/4
Dấu = xảy ra khi y=1/2
c: P=x^2-4x+4+y^2+y+1/4+11/4
=(x-2)^2+(y+1/2)^2+11/4>=11/4
Dấu = xảy ra khi x=2 và y=-1/2