tìm Max
A=5-8x-x^2
Bài 1)tìm Min hay Max
a) G=\(\dfrac{2}{x^2+8}\)
b) H=\(\dfrac{-3}{x^2-5x+1}\)
Bài 2) Tìm Min hay Max
a)D=\(\dfrac{2x^2-16x+41}{x^2-8x+22}\)
b)E=\(\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}\)
c)G=\(\dfrac{3x^2-12x+10}{x^2-4x+5}\)
1.
\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)
\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)
\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max
2.
\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)
\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)
\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)
\(E_{min}=-1\) khi \(x=0\)
\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)
\(G_{min}=-2\) khi \(x=2\)
Giúp mình với
câu 7: Tìm max
a)\(\sqrt{x-5}\)+\(\sqrt{23-x}\)
b)\(\sqrt{x-3}\)+\(\sqrt{19-x}\)
Với mọi số thực không âm a, b ta luôn có:
\(\left(a-b\right)^2\ge0\Leftrightarrow2ab\le a^2+b^2\)
\(\Leftrightarrow a^2+2ab+b^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)
Áp dụng:
a.
\(\sqrt{x-5}+\sqrt{23-x}\le\sqrt{2\left(x-5+23-x\right)}=6\)
Dấu "=" xảy ra khi \(x=14\)
b.
\(\sqrt{x-3}+\sqrt{19-x}\le\sqrt{2\left(x-3+19-x\right)}=4\sqrt{2}\)
Dấu "=" xảy ra khi \(x=11\)
tìm maxA=x+y+z biết x+2y=21;2x+5z=51
Cho 2 số thực x,y saocho \(x^2+y^2=1\)
Tìm minA, maxA biết A=x+y
Giúp em với ạ
Tìm x để A max
A= -x2-5
\(A=-x^2-5\)
Vì \(-x^2\le0\Leftrightarrow-x^2-5\le-5\)
Vậy \(A_{max}=-5\Leftrightarrow x=0\)
a) Tính giá trị của B
\(B=\sqrt{5-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}+\sqrt{9-4\sqrt{5}}\)
b) Tìm MaxA= x+y+z biết x+3y=21 ; 2x+5z=51
Tìm MaxA = (x^2 - x + 1)/(x^2 + x + 1)
Ta có :
\(A=\dfrac{x^2-x+1}{x^2+x+1}=\dfrac{3x^2+3x+3-2x^2-4x-2}{x^2+x+1}=\dfrac{3\left(x^2+x+1\right)-2\left(x+1\right)^2}{x^2+x+1}=3-\dfrac{2\left(x+1\right)^2}{x^2+x+1}\)
Do: \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow3-\dfrac{2\left(x+1\right)^2}{x^2+x+1}\le3\)
Vậy \(MAX_A=3\) . Dấu \("="\) xảy ra khi \(x=-1\)
Cho x, y dương và\(x^3+y^3=1.\) Tìm maxA=x+y.
\(1=x^3+y^3=\frac{x^4}{x}+\frac{y^4}{y}\ge\frac{\left(x^2+y^2\right)^2}{x+y}\ge\frac{\frac{\left(x+y\right)^4}{4}}{x+y}=\frac{\left(x+y\right)^3}{4}\)
\(\Leftrightarrow\)\(x+y\le\sqrt[3]{4}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{\sqrt[3]{2}}\)
Cho \(x\ge0\)
A=\(\frac{1}{\sqrt{x^2-2\sqrt{x}+10}}\)
Tìm MaxA
\(A=\frac{1}{\sqrt{\left(\sqrt{x}-1\right)^2+9}}\le\frac{1}{3}\)
MaxA= 1/3 khi x =1
Cho x,y,z > 0. Tìm :
a) \(maxA=\sqrt{x^2+\frac{1}{y^2}}+\sqrt{y^2+\frac{1}{z^2}}+\sqrt{z^2+\frac{1}{x^2}}\left(ĐK:x+y+z=1\right)\)
b) \(maxB=\sqrt{x^2+\frac{1}{y^2}}+\sqrt{y^2+\frac{1}{x^2}}\left(ĐK:x+y\le1\right)\)
c) \(max,minC=2x+\sqrt{5-x^2}\)