Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
Ta có: \(M=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{6}-\frac{1}{11}=\frac{5}{66}\)
\(\Rightarrow M=\frac{5}{66}:\frac{1}{2}=\frac{5}{33}.\)
\(M=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}\)
\(M=\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}\)
\(M=\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+\frac{2}{9\cdot10}+\frac{2}{10\cdot11}\)
\(M=2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\right)\)
\(M=2\left(\frac{1}{6}-\frac{1}{11}\right)\)
\(M=2\cdot\frac{5}{66}\)
\(M=\frac{5}{33}\)
b
Q=\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{9900}\)
Rồi giải tương tự như câu a là được
M=\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5.\frac{99}{100}=\frac{99}{20}\)
\(=\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+...+\frac{4}{18x20}\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2x\left(\frac{1}{2}-\frac{1}{20}\right)\\ =2x\frac{9}{20}\\ =\frac{9}{10}\)
520 +7,5 x 4 = \(\frac{X+175}{5}\)+ 30
<=>520+30=\(\frac{x+175}{5}\)+30
<=>\(\frac{X+175}{5}\)+30 = 550
<=> \(\frac{X+175}{5}\)= 520
<=> X+175 = 2600
<=> X=2425
mik chưa lm đc bài 1
k mik nha
\(=\frac{2017\times\left(2018-1\right)}{2018\times2016+2018-2017}\)
\(=\frac{2017\times2017}{2018\times\left(2016+1\right)-2017}\)
\(=\frac{2017\times2017}{2018\times2017-2017}\)
\(=\frac{2017\times2017}{2017\times\left(2018-1\right)}=\frac{2017\times2017}{2017\times2017}=1\)
bằng 0/2017