Có 25 số

Ta có:

\(2^x+2^{x+1}+2^{x+2}=224\)

\(\Rightarrow2^x+2^x.2+2^x.2.2=224\)

\(\Rightarrow2^x\left(1+2+4\right)=224\)

\(\Rightarrow2^x=224\div\left(1+2+4\right)\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^x=2^5\Rightarrow x=5\)

Vậy \(x=5\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\)

Ta có:

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^{2016}}\)

Vậy \(A=1-\dfrac{1}{2^{2016}}\)

​ai **** mình mình cho 3 ****

Giải:

Vì vai trò \(a,b,c,d\) bình đẳng

Giả sử \(a\ge b\ge c\ge d\) khi đó:

\(S=\left|a-b\right|+\left|a-c\right|+\left|a-d\right|+\left|b-c\right|+\left|b-d\right|+\left|c-d\right|\)

\(=\left(a-b\right)+\left(a-c\right)+\left(a-d\right)+\left(b-c\right)+\left(b-d\right)+\left(c-d\right)\)

\(=\left(3a+b\right)-\left(c+3d\right)\)

Do \(c+3d\ge0\Rightarrow S\le3a+b\)

\(S=3a+b\) khi \(c=d=0,\) lúc đó \(a+b=1\)

Do \(a\le1\) ta có \(S=2a+\left(a+b\right)=2a+1\le2.1+1\)

Hay \(S\le3\)

Vậy \(Max_S=3\) khi \(\left(a,b,c,d\right)=\left(1;0;0;0\right)\) và các hoán vị của nó