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a. \(2x^2-4x+10=x^2-2x+1+x^2-2x+1+8=\left(x-1\right)^2+\left(x-1\right)^2+8=2\left(x-1\right)^2+8\)
Vì \(2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2+8\ge8\)
Vậy...
b. \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy..
c. \(2x^2-6x+5=x^2-4x+4+x^2-2x+1=\left(x-2\right)^2+\left(x-1\right)^2\)
Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}}\Rightarrow\left(x-2\right)^2+\left(x-1\right)^2\ge0\)
Vậy...
\(a,-x^2+6x-16\)
\(=-x^2+3x+3x-9-5\)
\(=-x\left(x-3\right)+3\left(x-3\right)-5\)
\(=\left(3-x\right)\left(x-3\right)-5\)
\(=-\left(x-3\right)^2-5\le-5\)=>Luôn âm
\(c,-1+x-x^2\)
\(=-x^2+x-1\)
\(=-\left(x^2-x+\frac{1}{2}+\frac{1}{2}\right)\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\le\frac{-1}{2}\)=>Luôn âm
a,\(-\left(x^2-3x+4\right)\)
\(-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\Leftrightarrow-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)(luôn âm)
b\(-2\left(x^2-5x+\frac{15}{2}\right)\)
\(-2\left[\left(x-\frac{5}{2}\right)^2+\frac{5}{4}\right]\)
\(-2\left(x-\frac{5}{4}\right)^2-\frac{5}{2}\le-\frac{5}{2}\)(luôn âm)
c,\(-\left[\left(4x^2-4x+1\right)+\left(2y^2-6y+5\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y^2-3y+\frac{5}{2}\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2+\frac{1}{4}\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2\right]-\frac{1}{4}\le-\frac{1}{4}\)(luôn âm)
\(a;x^2-3x+3=x^2-2\cdot\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+3\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\Leftrightarrow x^2-3x+3>0\forall x\)
a.A= \(x^2+10x+27\)
\(=x^2+2.x.5+25+2\)
\(\left(x+5\right)^2+2\ge2\forall x\)
Dấu " = " xảy ra <=> x + 5 = 0
=> x = -5
Vậy Min A = 2 <=> x = -5
b.B = \(x^2-12x+37\)
\(=x^2-2.x.6+36+1\)
\(=\left(x-6\right)^2+1\ge1\forall x\)
Dấu " = " xảy ra <=> x - 6 = 0
=> x = 6
Vậy Min B = 1 <=> x = 6
c. \(x^2+x+7\)
\(=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{27}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\forall x\)
Dấu " =" xảy ra <=> \(x+\dfrac{1}{2}=0\)
\(x=\dfrac{-1}{2}\)
Vậy Min C = \(\dfrac{27}{4}\Leftrightarrow x=\dfrac{-1}{2}\)
Bài 1:
a) \(M=x^2-3x+10=\left(x^2-3x+\frac{9}{4}\right)+\frac{31}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
KL:...
2. a. \(A=12a-4a^2+3=-4\left(a-\frac{3}{2}\right)^2+12\)
Vì \(\left(a-\frac{3}{2}\right)^2\ge0\forall a\)\(\Rightarrow-4\left(a-\frac{3}{2}\right)^2+3\le3\)
Dấu "=" xảy ra \(\Leftrightarrow-4\left(a-\frac{3}{2}\right)^2=0\Leftrightarrow a-\frac{3}{2}=0\Leftrightarrow a=\frac{3}{2}\)
Vậy Amax = 3 <=> a = 3/2
b. \(B=4t-8v-v^2-t^2+2017=-\left(v^2+t^2-4t+8v+20\right)+2037\)
\(=-\left(t-2\right)^2-\left(v+4\right)^2+2037\)
Vì \(\left(t-2\right)^2\ge0;\left(v+4\right)^2\ge0\forall t;v\)
\(\Rightarrow-\left(t-2\right)^2-\left(v+4\right)^2+2037\le2037\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left(t-2\right)^2=0\\\left(v+4\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t-2=0\\v+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=2\\v=-4\end{cases}}\)
Vậy Bmax = 2037 <=> t = 2 ; v = - 4
c. \(C=m-\frac{m^2}{4}=-\frac{1}{4}\left(m-2\right)^2+1\)
Vì \(\left(m-2\right)^2\ge0\forall m\)\(\Rightarrow-\frac{1}{4}\left(m-2\right)^2+1\le1\)
Dấu "=" xảy ra \(\Leftrightarrow-\frac{1}{4}\left(m-2\right)^2=0\Leftrightarrow m-2=0\Leftrightarrow m=2\)
Vậy Cmax = 1 <=> m = 2
mk gợi ý, phần còn lại tự làm
a) \(A=x^2+2x+5=\left(x+1\right)^2+4\ge4\)
b) \(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
c) \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
d) \(D=x^2-2x+y^2-4y+7=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
e) \(E=x^2-4xy+5y^2+10x-22y+28=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
a) A = x2 + 2x + 5
= x2 + 2x + 1 + 4
= ( x + 1 )2 + 4
Nhận xét :
( x + 1 )2 > 0 với mọi x
=> ( x + 1 )2 + 4 > 4
=> A > 4
=> A min = 4
Dấu " = " xảy ra khi : ( x + 1 )2 = 0
=> x + 1 = 0
=> x = - 1
Vậy A min = 4 khi x = - 1
b) B = 4x2 + 4x + 11
= ( 2x )2 + 4x + 1 + 10
= ( 2x + 1 )2 + 10
Nhận xét :
( 2x + 1 )2 > 0 với mọi x
=> ( 2x + 1 )2 + 10 > 10
=> B > 10
=> B min = 10
Dấu " = " xảy ra khi : ( 2x + 1 )2 = 0
=> 2x + 1 = 0
=> x = \(\frac{-1}{2}\)
Vậy Bmin = 10 khi x = \(\frac{-1}{2}\)
c) C = ( x - 1 ) ( x + 2 ) ( x + 3 ) ( x + 6 )
= [ ( x - 1 ) ( x + 6 ) ] [ ( x + 2 ) ( x + 3 ) ]
= ( x2 + 5x - 6 ) ( x2 + 5x + 6 )
= ( x2 + 5x ) 2 - 62
= ( x2 + 5x )2 - 36
Nhận xét :
( x2 + 5x )2 > 0 với mọi x
=> ( x2 + 5x )2 - 36 > - 36
=> C > - 36
=> C min = - 36
Dấu " = " xảy ra khi : ( x2 + 5x )2 = 0
=> x2 + 5x = 0
=> x ( x + 5 ) = 0
=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy C min = - 36 khi x = 0 hoặc x = - 5
d) D = x2 - 2x + y2 - 4y + 7
= ( x2 - 2x + 1 ) + ( y2 - 4x + 4 ) + 2
= ( x - 1 )2 + ( y - 2 )2 + 2
Nhận xét :
( x - 1 )2 > 0 với mọi x
( y - 2 )2 > 0 với mọi y
=> ( x - 1 )2 + ( y - 2 )2 > 0
=> ( x - 1 )2 + ( y - 2 )2 + 2 > 2
=> D > 2
=> D min = 2
Dấu " = " xảy ra khi : \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)
=> \(\hept{\begin{cases}x-1=0\\y-2=0\end{cases}}\)
=> \(\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy D min = 2 khi x = 1 và y = 2
a, \(-x^2+2x-7=-\left(x^2-2x+1\right)+1-7=-\left(x-1\right)^2-6\)
Vì \(-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-6< 0\) => đpcm
b, \(-x^2-6x-10=-\left(x^2+6x+9\right)+9-10=-\left(x+3\right)^2-1\)
Vì \(-\left(x+3\right)^2\le0\Rightarrow-\left(x+3\right)^2-1< 0\) => đpcm
c, \(-x^2-3x-5=-\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{9}{4}-5=-\left(x+\dfrac{3}{2}\right)^2-\dfrac{11}{4}\)
Vì \(-\left(x+\dfrac{3}{2}\right)^2\le0\Rightarrow-\left(x+\dfrac{3}{2}\right)^2-\dfrac{11}{4}< 0\) => đpcm
d, \(-x^2+4xy-5y^2-8y-18=-\left(x^2-4xy+5y^2+8y+18\right)\)
\(=-\left[\left(x^2-4xy+4y^2\right)+\left(y^2+8y+16\right)\right]-1\)
\(=-\left[\left(x-2y\right)^2+\left(y+4\right)^2\right]-1=-\left(x-2y\right)^2-\left(y+4\right)^2-1\)
Vì \(-\left(x-2y\right)^2\le0;-\left(y+4\right)^2\le0\Rightarrow-\left(x-2y\right)^2-\left(y+4\right)^2-1< 0\)
=> đpcm