B  , 6,67

mik nghi the

\(A=\left|x+\frac{2}{3}\right|+1\frac{2}{3}\)

\(A=\left|x+\frac{2}{3}\right|+\frac{5}{3}\)

Áp dungk KT \(\left|x\right|\ge0\)\(\forall\)\(x\)

BG :

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

\(\Leftrightarrow\)\(\left|x+\frac{2}{3}\right|+\frac{5}{3}\ge0+\frac{5}{3}\)\(\forall\)\(x\)

hay \(A\ge\frac{5}{3}\)\(\forall\)\(x\)

Dấu "=" xảy ra :

\(\Leftrightarrow\)\(\left|x+\frac{2}{3}\right|=0\)

\(\Leftrightarrow\)\(x+\frac{2}{3}=0\)

\(\Leftrightarrow\)\(x=-\frac{2}{3}\)

Vậy GTNN của \(A\) bằng \(\frac{5}{3}\)đạt được khi \(x=-\frac{2}{3}\)

phải cần cách làm cơ

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!!!!!!!!!!!!!!