Tìm GTNN hoặc GTLN của các biểu thức: A 4 x 2 4 x11
D x 2 2 x y 2 4 y 7
B 5 – 8 x – x 2
E = 5 – x 2 + 2x – 4y2 – 4y
C = 4x – x 2 .
F = (x 2 2x x 2 2x 2
nhanh giúp mik đc k ạ
K
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KN
1 tháng 4 2018
D = \(-\dfrac{5}{x^2-4x+7}\)
Vì: x2 - 4x + 7
= x2 - 4x + 4 + 3
= (x - 2)2 + 3 \(\ge\) 3 \(\forall\)x
\(\Rightarrow\) \(\dfrac{5}{\left(x-2\right)^2+3}\) \(\le\) \(\dfrac{5}{3}\) \(\forall\)x
\(\Rightarrow\) \(-\dfrac{5}{\left(x-2\right)^2+3}\)\(\ge\)-\(\dfrac{5}{3}\) \(\forall\)x
Dấu"=" xảy ra khi:
x - 2 = 0
\(\Rightarrow\) x = 2
Vậy.............
E = \(\dfrac{2x^2+4x+4}{x^2+2x+4}\)
Ta có:
\(\dfrac{2x^2+4x+4}{x^2+2x+4}\)
= \(\dfrac{2\left(x^2+2x+4\right)-4}{x^2+2x+4}\)
= 2 - \(\dfrac{4}{x^2+2x+4}\)
Vì:
x2 + 2x + 4
= x2 + 2x + 1 + 3
= (x + 1)2 + 3 \(\ge\) 3 \(\forall\)x
\(\Rightarrow\) \(\dfrac{4}{\left(x+1\right)^2+3}\) \(\le\) \(\dfrac{4}{3}\) \(\forall\)x
\(\Rightarrow\) 2 - \(\dfrac{4}{\left(x+1\right)^2+3}\) \(\le\) \(\dfrac{2}{3}\) \(\forall\)x
Dấu "=" xảy ra khi:
x + 1 = 0
\(\Rightarrow\) x = -1
Vậy...............
F = \(\dfrac{6x+8}{x^2+1}\)
= \(\dfrac{x^2+6x+9-x^2-1}{x^2+1}\)
= \(\dfrac{\left(x+3\right)^2-\left(x^2+1\right)}{x^2+1}\)
= \(\dfrac{\left(x+3\right)^2}{x^2+1}-1\) \(\ge\) -1 \(\forall\)x
Dấu "=" xảy ra khi:
(x + 3)2 = 0
\(\Rightarrow\) x + 3 = 0
\(\Rightarrow\) x = -3
Vậy.....................
22 tháng 9 2021
Bài 5:
a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)
\(minA=5\Leftrightarrow x=2\)
b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Bài 4:
a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(maxM=7\Leftrightarrow x=2\)
b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
MA
11 tháng 9 2016
a) \(A=x^2-2x+5\)
\(A=x^2-2x+1+4\)
\(A=\left(x-1\right)^2+4\)
Có: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\)
Dấu '=' xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
Vậy: \(Min_A=4\) tại \(x=1\)
b) \(B=x^2+x+1\)
\(B=x^2+x+\frac{1}{4}+\frac{3}{4}\)
\(B=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Có: \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu '=' xảy ra khi: \(\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
Vậy: \(Min_B=\frac{3}{4}\) tại \(x=-\frac{1}{2}\)
MA
11 tháng 9 2016
c) \(C=4x-x^2+3\)
\(C=-x^2+4x-4+8\)
\(C=8-\left(x^2-4x+4\right)\)
\(C=8-\left(x-2\right)^2\)
Có: \(\left(x-2\right)^2\ge0\Rightarrow8-\left(x-2\right)^2\le8\)
Dấu '=' xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)
Vậy: \(Max_C=8\) tại \(x=2\)
