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#)Giải :
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{10}\)
\(A=\frac{1}{10}\)
Bài 1:
Ta có:
\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)
\(\Leftrightarrow N< M\)
Vậy \(M>N.\)
Bài 2:
Ta có:
\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)
\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)
\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
\(\Leftrightarrow A>B\)
Vậy \(A>B.\)
Bài 3:
\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)
\(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)
\(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)
Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)
\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm
\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)
Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)
Bài 4:
\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)
Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)
\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)
\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)
Vậy \(\frac{1991.1999}{1995.1995}< 1.\)
bài 1
a,
32 + 68 :17 x 5 - 29
= 32 + 20 -29
= 52 - 29
= 23
b,
15 x 48 - 30 x 24 - 125
= 720 - 720 -125
= 0-125
a,
32 + 68 :17 x 5 - 29
= 32 + 20 -29
= 52 - 29
= 23
b,
15 x 48 - 30 x 24 - 125
= 720 - 720 -125
= 0-125
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
Đổi : \(9\frac{2}{3}\)m = \(\frac{29}{3}\)m
Chiều rộng khu vườn đó là : \(\frac{29}{3}\times\frac{3}{4}=\frac{29}{4}\)( m )
Chu vi khu vườn đó là : \(\left(\frac{29}{3}+\frac{29}{4}\right)\times2=\frac{203}{6}\)( m )
Diện tích khu vườn đó là : \(\frac{29}{3}\times\frac{29}{4}=\frac{841}{12}\)( m\(^2\))
Vậy chu vi khu vườn là : \(\frac{203}{6}\)m
diện tích khu vườn là : \(\frac{841}{12}\)m\(^2\)
Đây chỉ là ý kiến của mk thôi còn tùy bạn tham khảo nhe !
\(\frac{2}{3\times5}\times a+\frac{2}{5\times7}\times a+...+\frac{2}{13\times15}\times a=\frac{28}{15}\)
=> \(\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}\right)\times x=\frac{28}{15}\)
=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\times x=\frac{28}{15}\)
=> \(\left(\frac{1}{3}-\frac{1}{15}\right)\times x=\frac{28}{15}\)
=> \(\frac{4}{15}\times x=\frac{28}{15}\)
=> \(x=\frac{28}{15}:\frac{4}{15}\)
-> \(x=7\)
\(\frac{2}{3\times5}\times a+\frac{2}{5\times7}\times a+...+\frac{2}{13\times15}\times a=\frac{28}{15}\)
\(a\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}\right)=\frac{28}{15}\)
\(a\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{28}{15}\)
\(a\times\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{28}{15}\)
\(a\times\frac{4}{15}=\frac{28}{15}\)
\(a=\frac{28}{15}:\frac{4}{15}\)
\(a=\frac{28}{15}\times\frac{25}{4}\)
\(a=\frac{28}{4}=7\)
\(A=\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{90}\)
Tổng trên có số số hạng là: \(\left(90-32\right)\div1+1=59\)
\(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{90}\)
\(>\frac{1}{45}+\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\)
\(=\left(\frac{1}{90}+\frac{1}{90}\right)+\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\)
\(=\frac{60}{90}=\frac{2}{3}\)
Đoàn Đức Hà: Tại sao dòng số 4 phân số đầu tiên lại là \(\frac{1}{45}\)ạ?