\(P=...\)

\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)

\(=\frac{1}{99}-1=\frac{-98}{99}\)

\(M=...\)

\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)

\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)

\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)

\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)

1/4 . 2/6 . 3/8 . ... .30/62 .31/64 = 2^x

(1/2 . 1/2).(2/3 . 1/2).(3/4 . 1/2). ... .(30/31 . 1/2).(31/32 . 1/2) = 2^x

(1/2.1/2. ... .1/2).(1/2 . 2/3 . 3/4. ... .30/31 . 31/32) = 2^x

   (31 số 1/2) 

(1/2)^31.  = 2^x

=> 0=x+36

      x=0-36

      x=-36

Vậy x=-36

Theo mk nghĩ,mk làm đúng nha .Tk cho mk

Để mk sửa phần này một chút

\((\frac{1}{2})^{31}\cdot\frac{1\cdot2\cdot3.....30\cdot31}{2\cdot3\cdot4.....31\cdot32}=2^x\)

\(\frac{1^{31}}{2^{31}}\cdot\frac{1}{32}=2^x\)

\(\frac{1}{2^{31}}\cdot\frac{1}{2^5}=2^x\)

\(\frac{1}{2^{36}}=2^x\)

\(1=2^x\cdot2^{36}\)

\(2^0=2^x+36\)

Rồi bn tự suy luận nha

\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.........................\frac{31}{64}=\frac{1}{2^x}\)

Có 31 thừa số

\(\Leftrightarrow\left(\frac{1}{2}.\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}\right)...................\left(\frac{1}{2}.\frac{31}{32}\right)=\frac{1}{2^x}\)

\(\Leftrightarrow\left(\frac{1}{2}.\frac{1}{2}...............\frac{1}{2}\right).\left(\frac{1}{2}.\frac{2}{3}...............\frac{31}{32}\right)=\frac{1}{2^x}\)

Có 31 thừa số                                            Có 31 thừa số

\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1.2....................31}{2.3..............32}=\frac{1}{2^x}\)

\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1}{32}=\frac{1}{2^x}\)

\(\Leftrightarrow\frac{1}{2^{31}}.\frac{1}{2^5}=\frac{1}{2^x}\)

\(\Leftrightarrow\frac{1}{2^{36}}=\frac{1}{2^x}\)

\(\Leftrightarrow x=36\)

Vậy x=36

Chúc bạn học tốt

a) \(\frac{1}{9}\)

b) -1100

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)

\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)

\(3A=1-\frac{1}{64}\)

\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)

a ) \(\left(-\frac{40}{51}.0,32.\frac{17}{20}\right):\frac{64}{75}\)

\(=\left(-\frac{40}{51}.\frac{8}{25}.\frac{17}{20}\right):\frac{64}{75}\)

\(=\left(\frac{-40.8.17}{51.25.20}\right):\frac{64}{75}\)

\(=\left(\frac{-16}{75}\right).\frac{75}{64}\)

\(=\frac{-1}{1}.\frac{1}{4}=-\frac{1}{4}\)

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