Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
So sánh các số sau đây:
a) 913,2324m3 < 913 232 413cm3.
b) \(\dfrac{12345}{1000}\)m3 = 12,345m3.
c) \(\dfrac{8372361}{100}\)m3 > 8 372 361dm3.
\(4\frac{2}{10}m=\frac{42}{10}m\)
\(3\frac{12}{100}m=\frac{312}{100}m\)
\(2\frac{2}{25}kg=2\frac{8}{100}kg=\frac{208}{100}kg\)
\(4\frac{2}{10}m=4,2m\)
\(3\frac{12}{100}m=3,12m\)
\(2\frac{2}{25}kg=2,08kg\)
nhấn vào đây: Câu hỏi của Hatsune Miku - Toán lớp 1 - Học toán với OnlineMath
câu của bn lun đó chắc thế! 54566955575675665665658768693534365465476767657558553
\(2^{100^3}=2^{1000000};3^{100^2}=3^{10000}\)
Vì 21000000 > 310000 nên \(2^{100^3}>3^{100^2}\)
\(\frac{M+5}{M+7}=\frac{M+7-2}{M+7}=1-\frac{2}{M+7}\)
\(\frac{M+2005}{M+2007}=\frac{M+2007-2}{M+2007}=1-\frac{2}{M+2007}\)
vì \(\frac{2}{M+7}>\frac{2}{M+2007}\Rightarrow\frac{M+5}{M+7}< \frac{M+2005}{M+2007}\)
a) \(\frac{5}{6}\)= \(\frac{15}{18}\); b) \(\frac{99}{100}\)< \(\frac{100}{99}\); c ) \(\frac{15}{17}\)> \(\frac{13}{18}\)vì \(\frac{15}{17}\)> \(\frac{15}{18}\)> \(\frac{13}{18}\);
d) \(\frac{222}{333}\)= \(\frac{2}{3}\)\(=1-\frac{1}{3}\); \(\frac{3333}{4444}\)= \(\frac{3}{4}\)= \(1-\frac{1}{4}\); vì \(\frac{1}{3}\)> \(\frac{1}{4}\)nên \(\frac{222}{333}\)< \(\frac{3333}{4444}\)
e) \(\frac{292929}{272727}\)= \(\frac{29}{27}\)= \(1+\frac{2}{17}\); \(\frac{347347}{345345}\)= \(\frac{347}{345}\)= \(1+\frac{2}{345}\)nên \(\frac{292929}{272727}\)> \(\frac{347347}{345345}\)
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 \(\frac{1991.1999}{1995.1995}< 1.\)
8/9 và 9/8 :
vì 8/9 < 1 và 9/8 > 1 nên 8/9 < 9/8
4/3 và 4/8 :
vì 3 < 8 nên 4/3 > 4/8
12/3 và 3/4 :
vì 12/3 > 1 và 3/4 nhỏ hơn 1 nên 12/3 > 3/4
mình giải chi tiết đó . nhớ l-i-k-e cho mk nha !
8372361 100 m 3 = 83723610d m 3 > 8372361d m 3