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a) đặt \(A=x^2+x+1\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=' xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=-\dfrac{1}{2}\)
b) đặt \(B=2+x-x^2\)
\(=-x^2+x+2\)
\(=-\left(x^2-x-2\right)\)
\(=-\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}-2\right]\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)
Vậy \(MAX_B=\dfrac{9}{4}\) khi \(x=\dfrac{1}{2}\)
c) đặt \(C=x^2-4x+1\)
\(=x^2-2\cdot x\cdot2+2^2-4+1\)
\(=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi \(x=2\)
Vậy \(MIN_c=-3\) khi \(x=2\)
d) đặt \(D=4x^2+4x+11\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2-1+11\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(MIN_D=10\) khi \(x=-\dfrac{1}{2}\)
mấy câu còn lại tương tự
\(A=x^2+12x+36=x^2+12x+36+3=\left(x+6\right)^2+3\ge3\)
Dấu '=' xảy ra khi x=-6
\(B=9x^2-12x+4-4=\left(3x-2\right)^2-4\ge-4\)
Dấu '=' xảy ra khi x=2/3
\(C=-x^2+4x+1\)
\(=-\left(x^2-4x-1\right)=-\left(x^2-4x+4-5\right)\)
\(=-\left(x-2\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=2
a) \(A=\left(x+1\right)\left(2x-1\right)\)
\(A=2x^2+2x-x-1\)
\(A=2x^2+x-1\)
\(A=2\left(x^2+\dfrac{1}{2}x-\dfrac{1}{2}\right)\)
\(A=2\left(x^2+2.x\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{2}\right)\)
\(A=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\)
Vì \(2\left(x+\dfrac{1}{4}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)
\(\Rightarrow Amin=-\dfrac{9}{8}\Leftrightarrow x=-\dfrac{1}{4}\)
\(B=4x^2-4xy+2y^2+1\)
\(B=\left(2x\right)^2-2.2x.y+y^2+y^2+1\)
\(B=\left(2x-y\right)^2+y^2+1\)
Vì \(\left(2x-y\right)^2\ge0\) với mọi x và y
\(y^2\ge0\) với mọi y
\(\Rightarrow\left(2x-y\right)^2+y^2+1\ge1\)
\(\Rightarrow Bmin=1\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
\(C=5x-3x^2+2\)
\(C=-\left(3x^2-5x-2\right)\)
\(C=-3\left(x^2-\dfrac{5}{3}x-\dfrac{2}{3}\right)\)
\(C=-3\left(x^2-2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{25}{36}-\dfrac{2}{3}\right)\)
\(C=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\)
Vì \(-3\left(x-\dfrac{5}{6}\right)^2\le0\) với mọi x
\(\Rightarrow-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\le\dfrac{49}{12}\)
\(\Rightarrow Cmax=\dfrac{49}{12}\Leftrightarrow x=\dfrac{5}{6}\)
\(D=-8x^2+4xy-y^2+3\)
\(D=-\left(4x^2-4xy+y^2\right)-4x^2+3\)
\(D=-\left(2x-y\right)^2-4x^2+3\)
Vì \(-\left(2x-y\right)^2\le0\) với mọi x và y
\(-4x^2\le0\) với mọi x
\(\Rightarrow-\left(2x-y\right)^2-4x^2+3\le3\) với mọi x và y
\(\Rightarrow Dmax=3\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
\(E=x^2-8x+38\)
\(E=x^2-2.x.4+16+22\)
\(E=\left(x-4\right)^2+22\)
Vì \(\left(x-4\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-4\right)^2+22\ge22\) với mọi x
\(\Rightarrow Emin=22\Leftrightarrow x=4\)
\(F=6x-x^2+1\)
\(F=-\left(x^2-6x-1\right)\)
\(F=-\left(x^2-2.x.3+9-9-1\right)\)
\(F=-\left(x-3\right)^2+10\)
Vì \(-\left(x-3\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x-3\right)^2+10\le10\)
\(\Rightarrow Fmax=10\Leftrightarrow x=3\)
\(a,x^2+2x+7\)
\(=x^2+2x+1+6\)
\(=\left(x+1\right)^2+6\)
\(V\text{ì}\left(x+1\right)^2\ge0\)
\(\left(x+1\right)^2+6\ge0+6\)
\(\left(x+1\right)^2+6\ge6\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\)
\(x+1=0\)
\(x=-1\)
Vậy MinA=6 khi x=-1
b) \(x^2+x+1\)
\(=x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\)
\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)
\(x=\dfrac{1}{2}\)
c, C= 4x^2 -12x +25
= 4x^2 -12x + 9+16
= (2x -3)^2 +16
ta có (2x-3)^2 >,= 0 với mọi x
=> (2x-3)^2 +16 >,=16 với mọi x
dấu bằng xảy ra khi (2x-3) ^2 =0
=> 2x-3 = 0
=> 2x =3
=> x =1,5
vậy .............
d, D = 2x^2 -8x -5
D= 2(x^2 -4x +4) -13
D= 2(x-2)^2 -13
ta có 2 (x-2)^2 >,= 0 với mọi x
=> 2(x-2)^2 -13 >,= -13 với mọi x
dấu = xảy ra khi 2(x-2)^2 =0
=> (x-2)^2=0
=>x-2 =0
=> x=2
vậy .............