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B) Ta có : \(1-\frac{1998}{1999}=\frac{1}{1999};1-\frac{1999}{2000}=\frac{1}{2000}\)
Vì 1999 < 2000 nên \(\frac{1}{1999}>\frac{1}{2000}\)
Hay \(\frac{1998}{1999}>\frac{1999}{2000}\)
A) Ta có : \(1-\frac{13}{27}=\frac{14}{27};1-\frac{27}{41}=\frac{14}{41}\)
Vì 27 < 41 nên \(\frac{1}{27}>\frac{1}{41}\)
Hay \(\frac{13}{27}>\frac{27}{41}\)
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a. Ta tính trước số bị chia: 1 + 4 + 7 + …… + 100
Dãy số gồm có: (100 – 1) : 3 + 1 = 34 (số hạng)
Ta thấy: 1 + 100 = 4 + 97 = 101 = …..
Do đó số bị chia là: 101 x 34 : 2 = 1717
Ta có: 1717 : a = 17
a = 1717 : 17
a = 101
vậy a = 101.
b.
x - 1 2 × 5 3 = 7 4 - 1 2 x - 1 2 × 5 3 = 5 4 x - 1 2 = 5 4 : 5 3 x - 1 2 = 3 4 x = 3 4 + 1 2 x = 5 4
c. 2000 2001 v à 2001 2002
Ta có: 1 - 2000 2001 = 1 2001
1 - 2001 2002 = 1 2002
Vì 1 2001 > 1 2002 nên 2000 2001 < 2001 2002
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\(vì\hept{\begin{cases}-18>-23\\91< 144\end{cases}}\Rightarrow\frac{-18}{91}>\frac{-23}{144}\)
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\(\frac{3737}{4747}\)\(=\frac{37}{47}\)
\(\frac{3737:11=37}{4747:11=47}\)
Hok tốt
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Câu 2
A= 1991 x1999= 1991 x(1995 + 4) = 1991 x1995 + 1991 x 4
B=1995x 1995= 1995 x (1991 + 4) = 1995 x 1991 + 1995 x 4
vì 1995 x 4 > 1991 x 4 nên 1995 x1991 + 1995 x 4 > 1991 x1995 + 1991 x 4 vậy A <B
M N P H O I K Q
\(a,\)* Xét hai tam giác MNK và KNP có :
+ Ta có : \(KM=\frac{1}{2}KP\)
+ Chung chiều cao hạ từ N
+ Do đó \(S_{MNK}=\frac{1}{2}S_{KNP}\)
b, Xét hai tam giác IKN và MNK có :
Ta có : \(IN=\frac{2}{3}MN\)
+ Chung chiều cao hạ từ K
+ Do đó : \(S_{IKN}=\frac{2}{3}S_{MNK}\)
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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 \(\frac{1991.1999}{1995.1995}< 1.\)
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1212/1313=12/13
2424/2525=24/25
phần bù của 12/13 là:1-12/13=1/13
phần bù của 24/25: 1-24/25=1/25
vì phần bù 1/13>1/25 nên 1212/1313>2424/2525