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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) Ta có A = x2 - 2x - 1 = (x2 - 2x + 1) - 2 = (x - 1)2 - 2 \(\ge\) -2
Dấu "=" xảy ra <=> x - 1 = 0 => x = 1
Vậy Min A = -2 <=> x = 1
b) Ta có B = 4x2 + 4x + 8 = (4x2 + 4x + 1) + 7 = (2x + 1)2 + 7 \(\ge\)7
Dấu |"=" xảy ra <=> 2x + 1 = 0 => x = -1/2
Vậy Min B = 7 <=> x = -1/2
c) Ta có C = 3x - x2 + 2
= -(x2 - 3x - 2)
= -(x2 - 3x + 9/4 - 9/4 - 2)
= -[(x - 3/2)2 - 17/4)
= -(x - 3/2)2 + 17/4 \(\le\frac{17}{4}\)
Dấu "=" xảy ra <=> x - 3/2 = 0 => x = 3/2
Vậy Max C = 17/4 <=> x = 3/2
d) Ta có D = -x2 - 5x = -(x2 + 5x) = -(x2 + 5x + 25/4 - 25/4) = -(x + 5/2)2 + 25/4 \(\ge\frac{25}{4}\)
Dấu "=" xảy ra <=> x + 5/2 = 0 => x = -5/2
Vậy Max D = 25/4 <=> x = -5/2
e) Ta có E = x2 - 4xy + 5y2 + 10x - 22y + 28
= (x2 - 4xy + 4y2) + 10x - 20y + y2 - 2y + 28
= (x - 2y)2 + 10(x - 2y) + 25 + (y2 - 2y + 1) + 2
= (x - 2y + 5) + (y - 1)2 + 2 \(\ge\)2
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vậy Min E = 2 <=> x = -3 ; y = 1
\(A=x^2-2x-1=x^2-2x+1-2=\left(x-1\right)^2-2\ge-2\)
Dấu \(=\)xảy ra khi \(x=1\). Vậy GTNN của \(A\)là \(-2\).
\(B=4x^2+4x+8=4x^2+4x+1+7=\left(2x+1\right)^2+7\ge7\)
Dấu \(=\)xảy ra khi \(x=\frac{-1}{2}\). Vậy GTNN của \(B\)là \(7\).
\(C=-x^2+3x+2=-x^2+2.\frac{3}{2}x-\left(\frac{3}{2}\right)^2+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\)
Dấu \(=\) xảy ra khi \(x=\frac{3}{2}\). Vậy GTLN của \(C\)là \(\frac{17}{4}\).
\(D=-x^2-5x=-x^2-2.\frac{5}{2}x-\left(\frac{5}{2}\right)^2+\frac{25}{4}=-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Dấu \(=\)xảy ra khi \(x=\frac{-5}{2}\). Vậy GTLN của \(D\) là \(\frac{25}{4}\).
\(E=x^2-4xy+5y^2+10x-22y+28\)
\(=x^2+4y^2+25-4xy+10x-20y+y^2-2y+1+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
Dấu \(=\)xảy ra khi \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\). Vậy GTNN của \(E\) là \(2\).
\(C=2\left(x-\frac{5}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\Rightarrow C_{min}=\frac{7}{8}\)
\(D=\left(x^2+4xy+4y^2\right)+\left(y^2+y+\frac{1}{4}\right)+\frac{8083}{4}\)
\(D=\left(x+2y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{8083}{4}\ge\frac{8083}{4}\)
\(E=\frac{1}{2}\left(4x^2+y^2+\frac{9}{4}-4xy-6x+3y\right)+\frac{1}{2}\left(y^2+y+\frac{1}{4}\right)+\frac{15}{4}\)
\(E=\frac{1}{2}\left(2x-y-\frac{3}{2}\right)^2+\frac{1}{2}\left(y+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\)
\(A=-\left(x-2\right)^2+11\le11\)
\(B=-\left(x+\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)
\(C=-\left(x-3y\right)^2-\left(y-2\right)^2+11\le11\)
3)
e)
b) Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0
1/
a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)
Dấu "=" xảy ra khi x=1/2
Vậy Amin=4 khi x=1/2
b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)
Dấu "=" xảy ra khi x=-1
Vậy Bmin = -4 khi x=-1
2/
a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)
Dấu "=" xảy ra khi x=3
Vậy Amax = 19 khi x=3
b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)
Dấu "=" xảy ra khi x=5/4
Vậy Bmax = 31/8 khi x=5/4
Bài 1
a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\frac{x}{2}-\frac{1}{2}\right)=2\left(x^2+2.\frac{1}{4}.x+\frac{1}{16}-\frac{9}{16}\right)\)\(=2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\)
Vì \(\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)
Dấu "=" xảy ra khi \(\left(x+\frac{1}{4}\right)^2=0\Leftrightarrow x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)
Vậy minA=-9/8 khi x=-1/4
b)\(B=4x^2-4xy+2y^2+1=\left(4x^2-4xy+y^2\right)+y^2+1=\left(2x-y\right)^2+y^2+1\)
Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)=>\(\left(2x-y\right)^2+y^2\ge0\Rightarrow B=\left(2x-y\right)^2+y^2+1\ge1\)
Dấu "=" xảy ra khi (2x-y)2=y2=0 <=> 2x-y=y=0 <=> x=y=0
Vậy minB=1 khi x=y=0
lý luận tương tự bài 1, bài này mình làm tắt
Bài 2:
a) \(C=5x-3x^2+2=-\left(3x^2-5x-2\right)=-3\left(x^2-\frac{5}{3}x-\frac{2}{3}\right)\)
\(=-3\left(x^2-2.\frac{5}{6}.x+\frac{25}{35}-\frac{49}{36}\right)=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{49}{36}\right]=\frac{49}{12}-3\left(x-\frac{5}{6}\right)^2\le\frac{49}{12}\)
Dấu "=" xảy ra khi x=5/6
b)\(D=-8x^2+4xy-y^2+3=3-\left(8x^2-4xy+y^2\right)=3-\left[\left(4x^2-4xy+y^2\right)+4x^2\right]\)
\(=3-\left[\left(2x-y\right)^2+4x^2\right]\le3\)
Dấu "=" xảy ra khi x=y=0