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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
Tìm GTLN:
\(A=-x^2+6x-15\)
\(=-\left(x^2-6x+15\right)\)
\(=-\left(x^2-2.x.3+9+6\right)\)
\(=-\left(x+3\right)^2-6\le0\forall x\)
Dấu = xảy ra khi:
\(x-3=0\Leftrightarrow x=3\)
Vậy Amax = - 6 tại x = 3
Tìm GTNN :
\(A=x^2-4x+7\)
\(=x^2+2.x.2+4+3\)
\(=\left(x+2\right)^2+3\ge0\forall x\)
Dấu = xảy ra khi:
\(x+2=0\Leftrightarrow x=-2\)
Vậy Amin = 3 tại x = - 2
Các câu còn lại làm tương tự nhé... :)
A=x2-4x+7
= x2-4x+4+3
= (x-2)2+3
Vì (x+2)2>/ 0
Nên (x-2)2+3>/3
Vậy MAX của A=3 khi x-2=0 => x=2
\(A=x^2-6x+3\)
\(=\left(x^2-6x+9\right)-6\)
\(=\left(x+3\right)^2-6\)
ma \(\left(x+3\right)^2\ge0\Leftrightarrow\left(x+3\right)^2-6\ge-6\)
vậy gtnn của A là -6 tại x=-3
\(B=x^2+3x+7=\left(x^2+2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{17}{4}\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\)
vay .............................................
2/
\(A=-x^2+4x+8=-\left(x^2-4x+4\right)+12=-\left(x-2\right)^2+12\le12\)
vay .........................................
\(B=-x^2+3x-5=-\left(x^2-2\frac{3}{2}x+\frac{9}{4}\right)-\frac{11}{4}=\left(x-\frac{3}{2}\right)^2-\frac{11}{4}\le-\frac{11}{4}\)
vay.....................................
nếu có sai mong bạn thông cảm
\(A=x^2-6x+10\)
\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)
\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\) \(\forall x\in z\)
\(\Leftrightarrow A_{min}=1khix=3\)
\(B=3x^2-12x+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\) \(\forall x\in z\)
\(\Leftrightarrow B_{min}=-11khix=2\)
a) \(A=x^2-3x-x+3+11\)
\(=\left(x^2-4x+4\right)+10\)
\(=\left(x-2\right)^2+10\ge10\forall x\in R\)
Dấu "=" xảy ra<=> \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
b) \(B=5-4x^2+4x\)
\(=-\left(4x^2-4x+1\right)+6\)
\(=-\left(2x-1\right)^2+6\le6\forall x\in R\)
Dấu "=" xảy ra<=> \(-\left(2x-1\right)^2=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
c) \(C=\left(x^2-3x+1\right)\left(x^2-3x-1\right)\)
\(=\left(x^2-3x\right)^2-1\ge-1\forall x\in R\)
Dấu "=" xảy ra<=>\(\left(x^2-3x\right)^2=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow x=0;x=3\)
1. a . 3x2 - 6x = 0
\(\Leftrightarrow3x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b. x3 - 13x = 0
\(\Leftrightarrow x\left(x^2-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)
c. 5x ( x - 2001 ) - x + 2001 = 0
<=> 5x ( x - 2001 ) - ( x - 2001 ) = 0
\(\Leftrightarrow\left(5x-1\right)\left(x-2001\right)=0\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2001=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2001\end{cases}}\)
a)A=4(x+11/8)^2 -153/16
Min A=-153/16 khi x=-11/8
b)B=3(x-1/3)^2 -4/3
Min B=-4/3 khi x=1/3
Bài 1:
a) \(A=4x^2+11x-2=\left(4x^2+11x+\dfrac{121}{16}\right)-\dfrac{153}{16}=\left(2x+\dfrac{11}{4}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\)
\(minA=-\dfrac{153}{16}\Leftrightarrow x=-\dfrac{11}{8}\)
b) \(B=3x^2-2x-1=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minB=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{3}\)
Bài 2:
a) \(A=-x^2+3x-1=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{5}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{3}{2}\)
b) \(B=-x^2-4x+7=-\left(x^2+4x+4\right)+11=-\left(x+2\right)^2+11\le11\)
\(maxB=11\Leftrightarrow x=-2\)