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\(P=...\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)
\(=\frac{1}{99}-1=\frac{-98}{99}\)
\(M=...\)
\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)
\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)
\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)
\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)
\(a.\)
\(A=\)\(\frac{10^{15}+1}{10^{16}+1}\)
\(10A=\) \(\frac{10\left(10^{15}+1\right)}{10^{16}+1}\)
\(10A=\) \(\frac{10^{16}+10}{10^{16}+1}\)
\(10A=\)\(\frac{10^{16}+1+9}{10^{16}+1}\)
\(10A=\frac{10^{16}+1}{10^{16}+1}+\frac{9}{10^{16}+1}\)
\(10A=1+\frac{9}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}\)
\(10B=\frac{10^{17}+1+9}{10^{17}+1}\)
\(10B=\frac{10^{17}+1}{10^{17}+1}+\frac{9}{10^{17}+1}\)
\(10B=1+\frac{9}{10^{17}+1}\)
\(\Rightarrow10B< 10A\Rightarrow B< A\)\(\text{( vì tự làm ) }\)
xin lỗi hôm qua mk đang làm thì phải đy học zoom học xong quên h mới nhơ ra làm típ :)
b
\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)
\(B=\frac{3}{8^4}+\frac{7}{8^3}=\frac{3}{8^4}+\frac{3}{8^3}+\frac{4}{8^3}\)
Vì \(\frac{4}{8^4}< \frac{4}{8^3}\)=.> A < B
a) Ta có\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{11}\right)=1-\frac{2}{11}=\frac{9}{11}\)
b) Ta có \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2048}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)(1)
Đặt S = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
Lấy 2S trừ S ta có :
2S - S \(=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\right)\)
\(S=1-\frac{1}{2048}\)
Khi đó (1) <=> \(1-\left(1-\frac{1}{2048}\right)=1-1+\frac{1}{2048}=\frac{1}{2048}\)
\(a,\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+....+\frac{2}{90}+\frac{2}{110}\)
\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.....+\frac{1}{90}+\frac{1}{110}\right)\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=1-\frac{2}{11}\)
\(=\frac{9}{11}\)
a)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
= \(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
b) Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-...-\frac{1}{2^6}+\frac{1}{2^6}-\frac{1}{2^7}\)
A =\(1-\frac{1}{2^7}\)
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=1-\frac{1}{11}\)
\(A=\frac{10}{11}\)
Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\left(1\right)\)
\(2B=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\frac{2}{2^5}+\frac{2}{2^6}+\frac{2}{2^7}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\left(2\right)\)
Lấy \(\left(2\right)-\left(1\right)\)hay \(2B-B\)ta có:
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Rightarrow B=1-\frac{1}{2^7}\)
\(\Rightarrow B=\frac{2^7-1}{2^7}=\frac{128-1}{128}=\frac{127}{128}\)
HOK TOT
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
Cứ quy đồng lần lượt thôi,có 6 phân số chứ mấy
tính nhanh bạn ạ.