A=(1+1/3+...+1/2019)-(1/2+1/4+...+1/2018)

A=(1+1/3+...+1/2019)+(1/2+1/4+...+1/2018)-(1/2+1/4+...+1/2018).2

A=(1+1/2+1/3+1/4+...+1/2019)-(1+1/2+...+1/1009)

A=1/1010+1/1011+...+1/2019

=) A=B

=) (A-B-1)^2019=-1

a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)

b, c cùng 1 câu phải k

ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)

\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)

A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)

NHA

HỌC TỐT

Đoán XemBạn lm đc chx ak

Ta có

\(A=\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\)                                                   \(B=\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)

\(\Leftrightarrow A=\frac{\left(\frac{17}{5}+\frac{1}{5}\right):\frac{5}{2}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{276}{56}}\)                                            \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}\)

\(\Leftrightarrow A=\frac{\frac{18}{5}:\frac{5}{2}}{\frac{89}{28}:\frac{276}{56}}\)                                                            \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(-\frac{1}{20}\right)}{\frac{2}{5}}\)

\(\Leftrightarrow A=\frac{\frac{36}{25}}{\frac{89}{138}}\)                                                                       \(\Leftrightarrow B=\frac{\frac{5}{4}}{\frac{2}{5}}\)

\(\Leftrightarrow A=\frac{4968}{2225}\)                                                                      \(\Leftrightarrow B=\frac{25}{8}\)

\(\Leftrightarrow A=\frac{39744}{17800}\)                                                                     \(\Leftrightarrow B=\frac{55625}{17800}\)

Ta có: 39744<55625

\(\Rightarrow A< B\)

Vậy A<B

kb vói mình đã

\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{2018}+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}=N\)

\(\Rightarrow M-N=0\Rightarrow\left(M-N\right)^2=0\)

best suarez    làm đúng rồi

Tại sao cái dòng thứ 2 là \(\frac{1}{3}\) mà xuống dòng dưới là \(\frac{1}{2}\)

A =\(\frac{\left(\frac{17}{5}+\frac{1}{5}\right).\frac{2}{5}}{\left(\frac{38}{7}-\frac{9}{4}\right).\frac{56}{267}}\)

A=\(\frac{36}{25}\).\(\frac{3}{2}\)=\(\frac{54}{25}\)=2,16

B=\(\frac{1,2:\left(\frac{6}{5}-\frac{5}{4}\right)}{0,32+\frac{2}{25}}\)=-24.\(\frac{5}{2}\)=-60

vì 2,16 > -60 Vậy A>B

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Leftrightarrow x+2=41\)

\(\Leftrightarrow x=41-2\)

\(\Leftrightarrow x=39\)

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99% LÀ 39

CÒN LAI LÀ ĐÁP ÁN KHÁC

b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)

d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)

Làm tiếp:

\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)

\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)

\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)

Bài 2:

Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)

\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)

\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)

\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)

Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)

Bài 1:Tính

a,   Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)

Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)

\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)

\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)

\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)

\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)

Áp dụng vào bài toán ta có đáp số là:1

b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)

c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)

d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)

e,Xét mẫu số ta có:

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)

\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)