\(P=1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2^{100}-1}=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+.......+\left(\frac{1}{2^{99}}+\frac{1}{2^{99}}+....+\frac{1}{2^{100}-1}\right)>1+\frac{2}{4}+\frac{4}{8}+.....+\frac{2^{99}}{2^{100}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+.....+\frac{1}{2}\left(101sohang\frac{1}{2}\right)=\frac{101}{2}>\frac{100}{2}=50\left(đpcm\right)\)

\(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=x-\left\{x-\left[x+x-1\right]\right\}=x-\left\{x-2x+1\right\}=x-\left\{-x+1\right\}=2x-1=1\Rightarrow2x=2\Rightarrow x=1.Vậy:x=1\)

\(M=2^0+2^1+2^2+2^3+2^4=1+2+4+8+16=31\left(\text{ cái này tính nguyên đi số nhỏ mà}\right)\)

số cần tìm là :

 0 x 1 = 0 ( vì bất kì số nào nhân với 0, cũng bằng 0)

vậy  sôs cần tìm là 0

\(\left\{{}\begin{matrix}\sqrt{\frac{x+3}{x}}+\frac{2y}{y-2}=8\left(1\right)\\2\sqrt{\frac{x+3}{x}}+\frac{3y}{y-2}=13\left(2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{2}\sqrt{\frac{x+3}{x}}+\frac{3y}{y-2}=12\\2\sqrt{\frac{x+3}{x}}+\frac{3y}{y-2}=13\end{matrix}\right..\text{Trừ vế theo vế ta được:}\frac{1}{2}\sqrt{\frac{x+3}{x}}=1\Leftrightarrow\sqrt{\frac{x+3}{x}}=2\Leftrightarrow\frac{x+3}{x}=4\Leftrightarrow x+3=4x\Leftrightarrow x=1.\text{ Thay x=1 vào}\left(1\right)taduoc:2+\frac{2y}{y-2}=8\Leftrightarrow\frac{2y}{y-2}=6\Leftrightarrow2y=6y-12\Leftrightarrow y=3\)

thử lại ta thấy t/m. Vậy: x=1;y=3