NN
2
K
Khách
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B
E =\(\frac{2018^{99}-1}{2018^{100}-1}\)so sánh với F =\(\frac{2018^{98}-1}{2018^{99}-1}\)
Ai nhanh tk
1
YN
26 tháng 2 2020
Ta có \(E=\frac{2018^{99}-1}{2018^{100}-1}\)
\(\Leftrightarrow2018E=\frac{2018^{100}-2018}{2018^{100}-1}\)
\(\Leftrightarrow2018E=1-\frac{2017}{2018^{100}-1}\) (2)
Lại có \(F=\frac{2018^{98}-1}{2018^{99}-1}\)
\(\Leftrightarrow2018F=\frac{2018^{99}-2018}{2018^{99}-1}\)
\(\Leftrightarrow2018F=1-\frac{2017}{2018^{99}-1}\) (2)
Mà \(2018^{100}>2018^{99}>0\)
\(\Leftrightarrow2018^{100}-1>2018^{99}-1\)
\(\Leftrightarrow\frac{2017}{2018^{100}-1}< \frac{2017}{2018^{99}-1}\)
\(\Leftrightarrow-\frac{2017}{2018^{100}-1}>-\frac{2017}{2018^{99}-1}\)
\(\Leftrightarrow1-\frac{2017}{2018-1}>1-\frac{2017}{2018^{99}-1}\) (3)
Từ (1) ;(2) và (3) <=> 2018E > 2018 F > 0
<=> E > F
Vậy E > F
@@ Học tốt
Chiyuki Fujito
K cần tk
DT
1
DT
2
10 tháng 2 2019
\(\frac{2018^{100}+1}{2018^{90}+1}\)= \(\frac{2018^{10}+1}{1+1}\)\(\frac{2018^{10}+1}{2}\)
\(\frac{2018^{99}+1}{2018^{89}+1}\)= \(\frac{2018^{10}+1}{1+1}\)= \(\frac{2018^{10}+1}{2}\)
=> \(\frac{2018^{100}+1}{2018^{90}+1}=\frac{2018^{99}+1}{2018^{89}+1}\)
nhớ bảo kê nha Duyên
NV
4 tháng 10 2018
\(2018^{100}+2018^{99}\)
\(=2018^{99}.\left(2018+1\right)\)
\(=2018^{99}.2019\)\(< 2019^{99}.2019=2019^{100}\)
\(\Rightarrow2018^{100}+2018^{99}< 2019^{100}\)
Vậy \(2018^{100}+2018^{99}< 2019^{100}\)
~~Hok tốt~~
21 tháng 6 2019
Bài toán : So sánh A và B
\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)
+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)
\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)
+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)
\(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)
.....................................
\(1=1\)
\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
JP
1
HD
5
TK
8 tháng 6 2018
Đặt : \(A=\frac{2018^{13}+1}{2018^{14}+1}\); \(B=\frac{2018^{2012}+1}{2018^{2013}+1}\)
Ta có :
\(2018A=\frac{2018.\left(2018^{13}+1\right)}{2018^{14}+1}\)
\(2018A=\frac{2018^{14}+2018}{2018^{14}+1}=\frac{2018^{14}+1+2017}{2018^{14}+1}=\frac{2018^{2014}+1}{2018^{14}+1}+\frac{2017}{2018^{14}+1}=1+\frac{2017}{2018^{14}+1}\)
\(2018B=\frac{2018.\left(2018^{12}+1\right)}{2018^{13}+1}\)
\(2018B=\frac{2018^{13}+2018}{2018^{13}+1}=\frac{2018^{13}+1+2017}{2018^{13}+1}=\frac{2018^{13}+1}{2018^{13}+1}+\frac{2017}{2018^{13}+1}=1+\frac{2017}{2018^{13}+1}\)
Vì 201814 + 1 > 201813 + 1 nên \(\frac{2017}{2018^{14}+1}< \frac{2017}{2018^{13}+1}\)
\(\Rightarrow1+\frac{2017}{2018^{14}+1}< 1+\frac{2017}{2018^{13}+1}\)Hay : A < B
Vậy A < B
IA
8 tháng 6 2018
Đặt \(A=\frac{2018^{13}+1}{2018^{14}+1}\)và \(B=\frac{2018^{12}+1}{2018^{13}+1}\)
Ta có :
\(2018A=\frac{\left(2018^{13}+1\right)\times2018}{2018^{14}+1}\) \(2018B=\frac{\left(2018^{12}+1\right)\times2018}{2018^{13}+1}\)
\(2018A=\frac{2018^{14}+2018}{2018^{14}+1}\) \(2018B=\frac{2018^{13}+2018}{2018^{13}+1}\)
\(2018A=\frac{2018^{14}+1+2017}{2018^{14}+1}\) \(2018B=\frac{2018^{13}+1+2017}{2018^{13}+1}\)
\(2018A=1+\frac{2017}{2018^{14}+1}\) \(2018B=1+\frac{2017}{2018^{13}+1}\)
Vì \(\frac{2017}{2018^{14}+1}< \frac{2017}{2018^{13}+1}\)
\(\Rightarrow2018A< 2018B\)
\(\Rightarrow A< B\)
Vậy : \(\frac{2018^{13}+1}{2018^{14}+1}< \frac{2018^{12}+1}{2018^{13}+1}\)
Áp dụng tính chất: Nếu \(\frac{a}{b}< 1\)thì \(\frac{a}{b}< \frac{a+m}{b+m}\)
Ta có: \(E=\frac{2018^{99}-1}{2018^{100}-1}< \frac{2018^{99}-1+2019}{2018^{100}-1+2019}\)
\(=\frac{2018^{99}-2018}{2018^{100}-2018}\)
\(=\frac{2018.\left(2018^{98}-1\right)}{2018.\left(2018^{99}-1\right)} \)
\(=\frac{2018^{98}-1}{2018^{99}-1}=F\)
Vậy \(E< F\)
Thanks bn