Ta có: \(\left|x+\frac{2}{3}\right|\ge0\forall x\)

\(\Rightarrow A\ge1\frac{3}{4}\forall x\)

Dấu '' = '' xảy ra khi: \(\left|x+\frac{2}{3}\right|=0\Rightarrow x=\frac{-2}{3}\)

Vậy \(MinA=1\frac{3}{4}\) khi \(x=\frac{-2}{3}\)

\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)

\(B=1-\frac{1}{2^{99}}< 1\)

Ta có : B = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

=> 2B = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

=> 2B - B = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)

=> \(B=1-\frac{1}{2^{99}}< 1\)

Đáp án là :  D  cả 3 số trên

nha!!!!!!!!!!!!!!

4627942781484

10¹² nha

5¹². 2¹²=(5.2)¹²=10¹²

Ta có:

\(2< 2,3< 3\Rightarrow\left[2,3\right]=2\)

\(0< \frac{1}{2}< 1\Rightarrow\left[\frac{1}{2}\right]=0\)

\(-4\le-4< -3\Rightarrow\left[-4\right]=-4\)

\(-6< 5,16< -5\Rightarrow\left[-5;16\right]=-6\)

+) 2 < 2,3 < 3

=> [ 2,3 ] = 2

+) \(0< \frac{1}{2}< 1\)

\(\Rightarrow\left[\frac{1}{2}\right]=0\)

+) \(-4\le-4< -3\)

\(\Rightarrow\left[-4\right]=-4\)

+) -6 < -5,16 < -5

=> [ - 5,16 ] = - 6