\(S=\dfrac{2018x^2-2.2018x+2018^2}{2018x^2}=\dfrac{2017x^2+x^2-2.2018x+2018^2}{2018x^2}=\dfrac{2017}{2018}+\dfrac{\left(x-2018\right)^2}{x^2}\ge\dfrac{2017}{2018}\)

\(S_{min}=\dfrac{2017}{2018}\) khi \(x=2018\)

c: \(=\left(x+1\right)^2+1>0\forall x\)

Trả lời:

a, \(x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTNN của biểu thức bằng 2 khi x = 3

b, \(-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-6x+9+2\right)=-\left[\left(x-3\right)^2+2\right]\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTLN của biểu thức bằng - 2 khi x = 3

c, \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\inℤ\)  (đpcm)

Dấu "=" xảy ra khi x + 1 = 0 <=> x = - 1

\(A=\dfrac{-x^2-2x+2014}{x^2}=\dfrac{2014}{x^2}-\dfrac{2}{x}-1=2014\left(\dfrac{1}{x}-\dfrac{1}{2014}\right)^2-\dfrac{2015}{2014}\ge-\dfrac{2015}{2014}\)

\(A_{min}=-\dfrac{2015}{2014}\) khi \(x=2014\)

sao mk ko nhìn thấy câu trả lời vậy bn

\(A=1-\frac{2}{x}+\frac{2014}{x^2}\)

đặt 1/x=t ta có

\(A=1-2t+2014t^2\)

   \(=2014\left(t^2-\frac{1}{1007}+\frac{1}{2014}\right)\)

   =\(2014[\left(t-\frac{1}{2014}\right)^2-\left(\frac{1}{2014}\right)^2+\frac{1}{2014}]\)

=\(2014\left(t-\frac{1}{2014}\right)^2+\frac{2013}{2014}\)\(\ge\frac{2013}{2014}\)

dấu''='' xảy ra khi t-1/2014=0 <=>1/x=1/2014=>x=2014

Ta có : \(A=\frac{x^2-2x+2014}{x^2}=\frac{2014x^2-4028x+2014^2}{x^2}=\frac{2013x^2+\left(x^2-4028x+2014^2\right)}{x^2}\)

\(=\frac{2013x^2}{x^2}+\frac{\left(x-2014\right)^2}{x^2}=2013+\frac{\left(x-2014\right)^2}{x^2}\)

Vì \(\frac{\left(x-2014\right)^2}{x^2}\ge0\forall x\)

Nên : \(A=2013+\frac{\left(x-2014\right)^2}{x^2}\ge2013\forall x\)

Vậy A_min = 2013 khi x = 2014

A=3(x^2+2/3x-1)

=3(x^2+2*x*1/3+1/9-10/9)

=3(x+1/3)^2-10/3>=-10/3

Dấu = xảy ra khi x=-1/3

\(B=1+\dfrac{15}{x^2+x+5}=1+\dfrac{15}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}}< =1+15:\dfrac{19}{4}=1+\dfrac{60}{19}=\dfrac{79}{19}\)

Dấu = xảy ra khi x=-1/2

b: Ta có: \(B=-2x^2+4x+1\)

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

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

\(=-2\left(x-1\right)^2+3\le3\forall x\)

Dấu '=' xảy ra khi x=1

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

1.

Đặt \(x-2=t\ne0\Rightarrow x=t+2\)

\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)

\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)

2.

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)

\(C_{max}=2\) khi \(t=3\) hay \(x=4\)