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1. a. \(A=8a-8a^2+3=-8\left(a-\frac{1}{2}\right)^2+5\)
Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow-8\left(a-\frac{1}{2}\right)^2+5\le5\)
Dấu "=" xảy ra \(\Leftrightarrow-8\left(a-\frac{1}{2}\right)^2=0\Leftrightarrow a-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{2}\)
Vậy Amax = 5 <=> a = 1/2
b. \(B=b-\frac{9b^2}{25}=-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\)
Vì \(\left(b-\frac{25}{18}\right)^2\ge0\forall b\)\(\Rightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\le\frac{25}{36}\)
Dấu "=" xảy ra \(\Leftrightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2=0\Leftrightarrow b-\frac{25}{18}=0\Leftrightarrow b=\frac{25}{18}\)
Vậy Bmax = 25/36 <=> b = 25/18
a,\(A=8a-8a^2+3\)
\(=-8\left(a^2-a\right)+3\)
\(=-8\left(a^2-2a\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)+3\)
\(=-8\left[\left(a-\frac{1}{2}\right)^2-\frac{1}{4}\right]+3\)
\(=-8\left(a-\frac{1}{2}\right)^2+2+3\)
\(=-8\left(a-\frac{1}{2}\right)^2+5\le5\forall a\)
Dấu"=" xảy ra khi \(\left(a-\frac{1}{2}\right)^2=0\Rightarrow a=\frac{1}{2}\)
Vậy \(Max_A=5\)khi\(a=\frac{1}{2}\)
bài 2:
b,\(D=d^2+10e^2-6de-10e+26\)
\(=d^2-23de+\left(3e\right)^2+e^2-2.5e+5^2+1\)
\(=\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\forall d,e\)
Dấu"=" xảy ra khi\(\orbr{\begin{cases}\left(d-3e\right)^2=0\\\left(e-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}d=15\\e=5\end{cases}}}\)
vậy \(D_{min}=1\)khi \(d=15;e=5\)
c,:\(E=4x^4+12x^2+11\)
\(=\left(2x^2\right)^2+2.2x^2.3+3^2+2\)
\(=\left(2x^2+3\right)^2+2\ge2\forall x\)
còn 1 đoạn nx bạn tự lm tiếp,lm giống như D
\(B=a^2\left(11-8a\right)+\left(2a-1\right)^3=11a^2-8a^3+\left(2a-1\right)^3=\left[\left(2a-1\right)^3-8a^3\right]+11a^2\)
\(=-12a^2+6a-1+11a^2=-a^2+6a-1=-\left(a^2-6a+9\right)+8=-\left(a-3\right)^2+8\)
Vậy giá trị lớn nhất của B là 8 tại a = 3
a) \(A=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\)
\(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+2\ge2\)
Đẳng thức xảy ra <=> x - 3 = 0 => x = 3
Vậy AMin = 2 , đạt được khi x = 3
b) \(B=5x-x^2=-x^2+5x=-x^2+5x-\frac{25}{4}+\frac{25}{4}=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
\(-\left(x-\frac{5}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2
Vậy BMax = 25/4 , đạt được khi x = 5/2
c) \(2x-2x^2-5=-2x^2+2x-5=-2\left(x^2-x+\frac{1}{4}\right)-\frac{9}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)
\(-2\left(x-\frac{1}{2}\right)^2\le0\forall x\Rightarrow-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le-\frac{9}{2}\)
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
Vậy CMax = -9/2 , đạt được khi x = 1/2
a)
\(B=4x^2+4x+2\)
\(=4x^2+4x+1+1\)
\(=\left(2x+1\right)^2+1\)
Nhận thấy: \(\left(2x+1\right)^2\ge0\)
=> \(\left(2x+1\right)^2+1>0\)
hay B luôn dương
a)
A=\(x^2+5x+7=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+7=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
C=\(3x^2-6x+5=\left[\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+\left(\sqrt{3}\right)^2\right]-\left(\sqrt{3}\right)^2+5\ge2 \)
b)
C=\(-x^2+4x-5=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\)
Ta có :\(\left(x-2\right)^2+1\ge1\Leftrightarrow-\left[\left(x-2\right)^2+1\right]\le\)-1
\(A=x^2+5x+7\)
\(A=\left(x^2+5x+\frac{25}{4}\right)+\frac{3}{4}\)
\(A=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x+\frac{5}{2}\right)^2=0\)
\(\Leftrightarrow\)\(x+\frac{5}{2}=0\)
\(\Leftrightarrow\)\(x=\frac{-5}{2}\)
Vậy GTNN của \(A\) là \(\frac{3}{4}\) khi \(x=\frac{-5}{2}\)
Chúc bạn học tốt ~
\(B=6x-x^2-5\)
\(-B=x^2-6x+5\)
\(-B=\left(x^2-6x+9\right)-4\)
\(-B=\left(x-3\right)^2-4\ge-4\)
\(B=-\left(x-3\right)^2+4\le4\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(x-3\right)^2=0\)
\(\Leftrightarrow\)\(x-3=0\)
\(\Leftrightarrow\)\(x=3\)
Vậy GTLN của \(B\) là \(4\) khi \(x=3\)
Chúc bạn học tốt ~
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
a,8a-8a2+3
=-8(a2-a)+3
=-8[a2-2a\(\dfrac{1}{2}\)+\(\left(\dfrac{1}{2}\right)^2\)-\(\dfrac{1}{4}\)]+3
=-8[(a-\(\dfrac{1}{2}\))2-\(\dfrac{1}{4}\)]+3
=-8(a-\(\dfrac{1}{2}\))2+2+3
=-8(a-\(\dfrac{1}{2}\))2+5
mà (a-\(\dfrac{1}{2}\))2\(\ge\)0
=>-8(a-\(\dfrac{1}{2}\))2\(\le\)0
=>-8(a-\(\dfrac{1}{2}\))2+5\(\le\)5
=> Gía trị lớn nhất biểu thức trên đạt được là 5( khi (a-\(\dfrac{1}{2}\))2=0\(\Leftrightarrow\)a=\(\dfrac{1}{2}\))