So sánh A=\(\frac{2017^{2017}}{1+2017+2017^2+....+2017^{2016}}\)
B=\(\frac{2016^{2017}}{1+2016+2016^2+...+2016^{2016}}\)
Những câu hỏi liên quan
So sánh A= \(\frac{2017^{2017}}{1+2017+2017^2+...+2017^{2016}}\)
B= \(\frac{2016^{2017}}{1+2016+2016^2+...+2016^{2016}}\)
SO SÁNH:
A = \(\frac{2017^{2017}}{1+2017+2017^2+...+2017^{2016}}\)
B = \(\frac{2016^{2017}}{1+2016+2016^2+...+2016^{2016}}\)
SO SÁNH A VÀ B:
A = \(\frac{2017^{2016}}{1+2017+2017^2+...+2017^{2016}}\)
B = \(\frac{2016^{2017}}{1+2016+2016^2+...+2016^{2016}}\)
so sánh A và B, biết:
A= \(\frac{2017^{2017}}{1+2017+2017^2+...+2017^{2016}}\) B= \(\frac{2016^{2017}}{1+2016+2016^2+...+2016^{2016}}\)
Câu1: So sánh:
A = \(\frac{2016^{2017}+2}{2016^{2017}-1}\) và B = \(\frac{2016^{2017}}{2016^{2017}-3}\)
Vì B > 1
=> \(B=\frac{2016^{2017}}{2016^{2017}-3}>\frac{2016^{2017}+2}{2016^{2017}-3+2}=\frac{2016^{2017}+2}{2016^{2017}-1}=A\)
Vậy A=B
Đúng 0
Bình luận (0)
so sanh A va B
\(A=\frac{2017^{100}}{1+2017+2017^2+2017^3+...+2017^{100}}\)
\(B=\frac{2016^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)
Ta có: \(A=\frac{2017^{100}}{1+2017+2017^2+2017^3+...+2017^{100}}\)
\(\Leftrightarrow A=\frac{\left[\left(20.100\right)+16+1\right]^{100}}{1+2017+2017^2+2017^3+...+2017^{10}}\)
\(B=\frac{2016^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)
\(\Leftrightarrow B=\frac{\left[\left(20.100+16\right)\right]^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)
Ta có hai tổng A và B mới để so sánh:
\(A=\frac{\left[\left(20.100\right)+16+1\right]^{100}}{1+2017+2017^2+2017^3+...+2017^{100}}\)
\(B=\frac{\left[\left(20.100\right)+16\right]^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)
Tới đây đơn giản rồi. Bạn làm tiếp đi nhé! Mẹ mình bắt tắt máy không cho làm nên đành dừng lại ở đây thôi! Thông cảm :V
Đúng 0
Bình luận (0)
So sánh 2 phân số
A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Đúng 0
Bình luận (0)
So sánh hai phân số : A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Ta có:
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Cộng vế theo vế, ta có:
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Vậy A > B
Đúng 0
Bình luận (0)
Bạn có nhầm không, tớ thấy cả hai đều giống nhau mà,
Hai cái bằng nhau
So sánh
\(A=\frac{2017^{2016}-2020}{2017^{2016}-6056}\)
\(B=\frac{2017^{2016}+2018}{2017^{2016}-2018}\)