Question

Tìm GTLN của: 

a, A= x^4 + 3x^2 + 2

b, B= 1/ 2(x-1)^2 + 3

c, C= x^2+8/ x^2+2

Answer 1

a) Vì x⁴ +3x² > Hoặc =0 Với mọi x

=> x⁴ +3x²+2 > Hoặc = 2 Với mọi x

Hay A > hoặc bằng 2 vs mọi x ..........

Answer 2

b)\(B=\frac{1}{2\left(x-1\right)^2+3}\)

Thấy: \(\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2\ge0\)

\(\Rightarrow2\left(x-1\right)^2+3\ge3\)

\(\Rightarrow\frac{1}{2\left(x-1\right)^2+3}\le\frac{1}{3}\)

Khi x=1

c)\(\frac{x^2+8}{x^2+2}=\frac{x^2+2+6}{x^2+2}=\frac{x^2+2}{x^2+2}+\frac{6}{x^2+2}=1+\frac{6}{x^2+2}\)

Thấy \(x^2\ge0\Rightarrow x^2+2\ge2\)

\(\Rightarrow\frac{1}{x^2+2}\le\frac{1}{2}\Rightarrow\frac{6}{x^2+2}\le\frac{6}{2}=3\)

\(\Rightarrow1+\frac{6}{x^2+2}\le1+3=4\)

Khi x=0

Answer 3

b)\(B=\frac{1}{2\left(x-1\right)^2+3}\)

Thấy: \(\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2\ge0\)

\(\Rightarrow2\left(x-1\right)^2+3\ge3\)

\(\Rightarrow\frac{1}{2\left(x-1\right)^2+3}\le\frac{1}{3}\)

Khi x=1

c)\(\frac{x^2+8}{x^2+2}=\frac{x^2+2+6}{x^2+2}=\frac{x^2+2}{x^2+2}+\frac{6}{x^2+2}=1+\frac{6}{x^2+2}\)

Thấy \(x^2\ge0\Rightarrow x^2+2\ge2\)

\(\Rightarrow\frac{1}{x^2+2}\le\frac{1}{2}\Rightarrow\frac{6}{x^2+2}\le\frac{6}{2}=3\)

\(\Rightarrow1+\frac{6}{x^2+2}\le1+3=4\)

Khi x=0

Answer 4

ban kia lam dung roi do

k tui nha

thanks