so sánh 2 phân số
A = \(\frac{2017x2018-1}{2017x2018-2}\)và B = \(\frac{2017}{2018}\)
K
Khách
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VT
2
4 tháng 7 2018
\(\frac{2016+2017.2018}{2017.2019-1}\)
= \(\frac{2016+2017.2018}{2017.2018+2017-1}\)
= \(\frac{2016+2017.2018}{2017.2018+2016}\)
= 1
LP
1
KV
30 tháng 10 2023
A = 1.2 + 2.3 + 3.4 + ... + 2017.2018
⇒ 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.218.(2019 - 2016)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2017.2018.2019 - 2016.2017.2018
= 2017.2018.2019
= 2017.2018.2019
B = 2018³/3 ⇒ 3B = 2018³
Ta có:
2017.2019 = (2018 - 1).(2018 + 1)
= 2018² - 1²
= 2018.2018 - 1 < 2018.2018
⇒ 2017.2018.2019 < 2018.2018.2018
⇒ 3A < 3B
⇒ A < B
3 tháng 12 2017
Ta có \(A=\frac{2017-2018}{2017+2018}=\frac{\left(2017-2018\right)\left(2017+2018\right)}{\left(2017+2018\right)^2}=\frac{2017^2-2018^2}{2017^2+2018^2+2.2017.2018}< \frac{2017^2-2018^2}{2017^2+2018^2}=B\)
Vậy A<B
PT
2
14 tháng 7 2017
Ta thấy \(A=\frac{2018-2017}{2018+2017}=\frac{2018^2-2017^2}{\left(2018+2017\right)^2}=\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}\)
Mà \(2018^2+2.2018.2017+2017^2>2018^2+2017^2\)
\(\Rightarrow\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}< \frac{2018^2-2017^2}{2018^2+2017^2}\)
Vậy A<B
DL
11 tháng 6 2018
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 \(\frac{1991.1999}{1995.1995}< 1.\)
5 tháng 5 2018
Ta có : \(B=\frac{2015+2016+2017}{2016+2017+2018}\) \(=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2016}\)
Cộng vế theo vế, ta có :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
M
3
31 tháng 8 2020
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(C=1-\frac{1}{2018}\)
\(C=\frac{2017}{2018}\)
31 tháng 8 2020
\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2017x2018}\)
Ta thấy \(\frac{1}{1x2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2x3}=\frac{1}{2}-\frac{1}{3}\)
.............................................
\(\frac{1}{2017x2018}=\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{1}{1}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{2017}{2018}\)
Chúc bạn học tốt nhớ k mình nhá
Ta có:
\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}\)
\(A=\frac{2017\cdot2018-2+1}{2017\cdot2018-2}\)
\(A=\frac{2017\cdot2018-2}{2017\cdot2018-2}+\frac{1}{2017\cdot2018-2}\)
\(A=1+\frac{1}{2017\cdot2018-2}\)
Ta có phân số trung gian là 1. Ta có:
\(A>1\) ; \(B< 1\)
\(\Rightarrow A>1>B\)
\(\Rightarrow A>B\)
Vậy A>B
Chúc em học tốt!
\(\Rightarrow\text{❤️✔✨♕✨✔️❤ }\Leftarrow\)
\(\text{Ta có :}\)
\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}=\frac{4070305}{4070304}=1\frac{1}{4070304}\)
\(B=\frac{2017}{2018}\)
\(\text{Vì : }1\frac{1}{4070304}>1\text{ mà }\frac{2017}{2018}< 1\text{ nên }1\frac{1}{4070304}>\frac{2017}{2018}\)
\(\Rightarrow A>B\)