CMR
A=\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+.......+\frac{1}{64}>1\)
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\(P=...\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)
\(=\frac{1}{99}-1=\frac{-98}{99}\)
\(M=...\)
\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)
\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)
\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)
\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)
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
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)
\(3A=1-\frac{1}{64}\)
\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)
a ) \(\left(-\frac{40}{51}.0,32.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(-\frac{40}{51}.\frac{8}{25}.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(\frac{-40.8.17}{51.25.20}\right):\frac{64}{75}\)
\(=\left(\frac{-16}{75}\right).\frac{75}{64}\)
\(=\frac{-1}{1}.\frac{1}{4}=-\frac{1}{4}\)