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b) B= 5x2 -10x+3-2
B = (5x2 - 2.5.1 . 12)-2
B = (5x-1)2-2
ta có :
(5x-1)2 > 0 với mọi x thuộc R
(5x-1)2 -2 < -2
vậy B < -2
dấu = xảy ra <=> x = 1/5
mai tui lm nốt choa
a)
\(A=4x^2-4x-1=4x^2-4x+1-2=\left(2x-1\right)^2-2\)
\(A\ge-2\forall x\in R\)
Dấu "=" xảy ra <=>\(\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
Vậy Amin =-2 tại x=1/2
Lời giải:
a)
\(A=4x^2-4x+1=2x(2x-3)+2x+1=2x(2x-3)+(2x-3)+4\)
\(=(2x+1)(2x-3)+4\)
Với \(x\geq \frac{3}{2}\Rightarrow \left\{\begin{matrix} 2x+1>0\\ 2x-3\geq 0\end{matrix}\right.\Rightarrow A=(2x+1)(2x-3)+4\geq 4\)
Vậy GTNN của $A$ là $4$ khi $x=\frac{3}{2}$
b)
\(B=5x^2-10x+3=5(x^2-2x+1)-2\)
\(=5(x-1)^2-2\)
Ta thấy \((x-1)^2\geq 0, \forall x\geq 1\Rightarrow B=5(x-1)^2-2\geq -2\)
Vậy GTNN của $B$ là $-2$ khi $(x-1)^2=0\Leftrightarrow x=1$
c)
\(C=4x^2-6x+2=(2x)^2-2.2x.\frac{3}{2}+(\frac{3}{2})^2-\frac{1}{4}\)
\(=(2x-\frac{3}{2})^2-\frac{1}{4}\)
Ta thấy \((2x-\frac{3}{2})^2\geq 0, \forall x\geq 0\Rightarrow C=(2x-\frac{3}{2})^2-\frac{1}{4}\geq -\frac{1}{4}\)
Vậy GTNN của $C$ là $\frac{-1}{4}$ khi \((2x-\frac{3}{2})^2=0\Leftrightarrow x=\frac{3}{4}\)
d)
\(D=3x^2+2x+1=3(x^2+\frac{2}{3}x+\frac{1}{9})+\frac{2}{3}\)
\(=3(x+\frac{1}{3})^2+\frac{2}{3}\)
Ta thấy \((x+\frac{1}{3})^2\geq 0, \forall x\geq -1\Rightarrow D=3(x+\frac{1}{3})^2+\frac{2}{3}\geq \frac{2}{3}\)
Vậy GTNN của $D$ là $\frac{2}{3}$ khi $(x+\frac{1}{3})^2=0\Leftrightarrow x=-\frac{1}{3}$
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
a, \(A=9x^2-6x+5\)
\(=\left(9x^2-6x+1\right)+4\)
\(=\left(3x-1\right)^2+4\)
ta có:
\(\left(3x-1\right)^2\ge0\forall x\Rightarrow\left(3x-1\right)^2+4\ge4\forall x\)
Vậy Min A = 4
Để A = 4 thì \(3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(b,B=4x^2-5x\)
\(=\left(4x^2-5x+\dfrac{25}{16}\right)-\dfrac{25}{16}\)
\(=\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\)
TA có:
\(\left(2x-\dfrac{5}{4}\right)^2\ge\forall x\Rightarrow\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\ge-\dfrac{25}{16}\forall x\)Vậy Min B = \(-\dfrac{25}{16}\)
Để B = \(-\dfrac{25}{16}\) thì \(2x-\dfrac{5}{4}=0\Rightarrow2x=\dfrac{5}{4}\Rightarrow x=\dfrac{5}{8}\)
\(c,C=3x^2-6x\)
\(=3\left(x^2-2x+1\right)-3\)
\(=3\left(x-1\right)^2-3\)
Ta có:
\(3\left(x-1\right)^2\ge0\forall x\Rightarrow3\left(x-1\right)^2-3\ge-3\)
vậy Min C = -3
Để C = -3 thì x-1=0 => x = 1
\(d,D=5x^2-15x\)
\(=5\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{45}{4}\)
\(=5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\)
Ta có:
\(5\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)Vậy Min D = \(-\dfrac{45}{4}\)
Để \(D=-\dfrac{45}{4}\) thì \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(e,E=x^2+3x+4\)
\(=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Vậy Min E = \(\dfrac{7}{4}\) khi \(x+\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(f,F=2x^2-4x+7\)
\(=2\left(x^2-2x+1\right)+5\)
\(=2\left(x-1\right)^2+5\ge5\forall x\)
Vậy Min F = 5 khi x - 1 =0 => x = 1
\(g,2x^2-3x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\)
\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)
Vậy Min G = \(\dfrac{-9}{8}\) khi \(x-\dfrac{3}{4}=0\Rightarrow x=\dfrac{3}{4}\)
\(h,H=3x^2-4x=3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{4}{3}\)
\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)
Vậy Min H = \(-\dfrac{4}{3}\) khi \(x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
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
\(x^2+y^2-xy-2x-2y+9=x^2+y^2+2xy-2x-2y+9-3xy\)
\(=\left(x+y\right)^2-2\left(x+y\right)+9-3xy=\left(x+y-2\right)\left(x+y\right)+9-3xy.\)
\(đếnđâytịt\)
b
c, =3 dễ
\(\frac{3x^2-6x+9}{x^2-2x+3}=\frac{3\left(x^2-2x+3\right)}{x^2-2x+3}=3\)