A=\(\frac{2017^{2016}+1}{2017^{2017}+1}\) B=\(\frac{2017^{2017}+1}{2017^{2018}+1}\)
So sánh : A=\(\frac{2017^{2016}+1}{2017^{2017}+1}\) B=\(\frac{2017^{2017}+1}{2017^{2018}+1}\)
so sánh A=\(\frac{2016^{2017}+1}{2017^{2018}+1}\)và B=\(\frac{2017^{2018}+1}{2017^{2017}+1}\)
Vì phân số A\(=\frac{2016^{2017}+1}{2017^{2018}+1}< 1\) mà B\(=\frac{2017^{2018}+1}{2017^{2017}+1}>1\)
\(\Rightarrow\frac{2016^{2017}+1}{2017^{2018}+1}< 1< \frac{2017^{2018}+1}{2017^{2017}+1}\)
Vậy A<B
Câu 1. Tính hợp lý giá trị các biểu thức sau :
a. A = ( 689 - 31 ) - ( 269 - 131 )
b. B = \(\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}+1\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}\right)-\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}+1\right)\)c. C = \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
tính
A=\(\frac{\frac{2017}{2}+\frac{2017}{3}+\frac{2017}{4}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)
\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)
\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
Giờ ta thế vào bài toán ban đầu được
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)
\(=\frac{2017}{2018}\)
\(A=\frac{1}{2017}+\frac{2}{2017^2}+\frac{3}{2017^3}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\). Chứng minh rằng : A < \(\frac{2017}{2016^2}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}\)
\(B=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\)
Tính \(\left(A^{2017}-B^{2017}\right)^{2018}\)
Ta có:
\(\Rightarrow A=B.\)
\(\Rightarrow A^{2017}=B^{2017}\)
\(\Rightarrow\left(A^{2017}-B^{2017}\right)^{2018}=\left(B^{2017}-B^{2017}\right)^{2018}=0^{2018}=0.\)
Vậy \(\left(A^{2017}-B^{2017}\right)^{2018}=0.\)
Chúc bạn học tốt!
tính :
\(A=\frac{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)
giúp mình đc ko mọi ng ???
ai nhanh nhất mình k cho nha.
Mình giúp bạn nha!
A = 2017/1 + 2017/2 + 2017/3 + . . . + 2017/2018 / 2017/1 + 2016/2 + 2015/3 + . . .+ 1/2017
= 2017 . ( 1 + 1/2 + 1/3 + . . . +1/2018 ) / ( 2017 . 2016 . 2015 . . . 1) . ( 1 + 1/2 + 1/3 +. . . + 1/2017 )
= 1/2016 . 2015 . 2014. . . 1
k mình nha
Trịnh Lê Na làm kiểu j mà chẳng hiểu thế
1.So sánh \(\frac{2016}{2017}+\frac{2017}{2018}\)với \(1\)( không tính kết quả )
2.So sánh: \(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
3. Với n là số nguyên dương hãy so sánh 2 phân số sau: \(\frac{n}{n+8}\)và \(\frac{n-2}{n+9}\)
1. \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)>1
2. A>B
A = \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)và B = \(\frac{2016+2017+2018}{2017+2018+2019}\)
\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)
\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)
\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)
\(\Rightarrow A\)>\(3-1=2\)
\(B=\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow B=1-\frac{3}{6054}\)
\(\Rightarrow B=1-\frac{1}{2018}\)
\(B\)<\(1\);\(A\)>\(2\)
\(\Rightarrow A\)>\(B\)