Question

[Trần Huệ]

Tìm GTLN của biểu thức:

M = \(\frac{x^2}{x^4+x^2+1}\)

giúp mik vs 

Answer 1

[Đinh quang hiệp]

\(x^4+x^2>=2\sqrt{x^4\cdot x^2}=2x^3;x^2+1>=2\sqrt{x^2}=2x;x^4+1>=2\sqrt{x^4}=2x^2\)(bđt cosi)

\(\Rightarrow x^4+x^2+x^2+1+x^4+1=2\left(x^4+x^2+1\right)>=2\left(x^3+x+x^2\right)\Rightarrow x^4+x^2+1>=x^3+x^2+x\)

\(\Rightarrow M=\frac{x^2}{x^4+x^2+1}< =\frac{x^2}{x^3+x^2+x}\)

\(x^3+x^2+x>=3\sqrt[3]{x^3x^2x}=3\sqrt[3]{x^6}=3x^2\)(bđt cosi)\(\Rightarrow\frac{x^2}{x^3+x^2+x}< =\frac{x^2}{3x^2}=\frac{1}{3}\Rightarrow M< =\frac{1}{3}\)

dáu = xảy ra khi x=1

vậy max M là \(\frac{1}{3}\)khi x=1

Answer 2

[Đinh quang hiệp]

mk lm sai rồi lm lại nhé

\(x^4,x^2>=0;1>0\Rightarrow x^4+x^2+1>=3\sqrt[3]{x^4\cdot x^2\cdot1}=3\sqrt[3]{x^6}=3x^2\)(bđt cosi)

\(\Rightarrow\frac{x^2}{x^4+x^2+1}< =\frac{x^2}{3x^2}=\frac{1}{3}\)

dấu = xảy ra khi \(x^4=x^2=1\Rightarrow x=+-1\)

vậy max M là \(\frac{1}{3}\)khi x=+-1