Question

Tìm giá trị lớn nhất của biểu thức sau: \(B=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}\)

Answer 1

\(B=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)

Do \(x^2\ge0\forall x\Rightarrow x^2+1\ge1\forall x\)

\(\Rightarrow B=\dfrac{3}{x^2+1}\le\dfrac{3}{1}=3\)

\(maxB=3\Leftrightarrow x^2=0\Leftrightarrow x=0\)

Answer 2

3x

Answer 3

\(B=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+x+1}=\dfrac{3}{x^2+1}\)

Ta có : \(x^2+1\ge1\Rightarrow\dfrac{3}{x^2+1}\le3\)

Dấu ''='' xảy ra khi x =0 

Vậy với x = 0 thì B đạt GTLN là 3

Answer 4

=\(\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\)

=\(\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}\)

=\(\dfrac{3}{x^2+1}\)

Ta có \(x^2\ge0\)

=>\(x^2+1\ge1\)

=>\(\dfrac{3}{x^2+1}\le3\)

hay \(B\le3\)

Dấu = xảy ra khi và chỉ khi x=0

Answer 5

\(B=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+x+1}\)

\(=\dfrac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}=\dfrac{3}{x^2+1}\) (đk: \(x\ne1\))

Vì \(\left(x^2+1\right)\ge1,\forall x\)

\(\Rightarrow B\le3\forall x\)

Min \(B=3\Leftrightarrow x=0\)