so sánh:
\(\frac{2^{2015}+1}{2^{2012}+1}\)và \(\frac{2^{2017}+1}{2^{2014}+1}\)
So sánh \(\frac{2^{2015}+1}{2^{2012}+1}\)và \(\frac{2^{2017}+1}{2^{2014}+1}\)
Giả sử A=\(\frac{2^{2015}+1}{2^{2012}+1}\)
-->\(\frac{1}{2^3}A=\frac{2^{2015}+1}{2^{2015}+8}\)
\(\frac{1}{8}A=\frac{2^{2015}+1}{2^{2015}+1}+\frac{2^{2015}+1}{7}\)
\(\frac{1}{8}A=1+\frac{2^{2015}+1}{7}\)
B=\(\frac{2^{2017}+1}{2^{2014}+1}\)
\(\frac{1}{2^3}B=\frac{2^{2017}+1}{2^{2017}+8}\)
\(\frac{1}{8}B=\frac{2^{2017}+1}{2^{2017}+1}+\frac{2^{2017}+1}{7}\)
\(\frac{1}{8}B=1+\frac{2^{2017}+1}{7}\)
Vì \(1+\frac{2^{2015}+1}{7}< 1+\frac{2^{2017}+1}{7}\)
nên \(\frac{1}{8}A< \frac{1}{8}B\)
-->A<B
-->\(\frac{2^{2015}+1}{2^{2012+1}}< \frac{2^{2017+1}}{2^{2014}+1}\)
\(\frac{2^{2015}+1}{2^{2012}+1}và\frac{2^{2017}+1}{2^{2014}+1}\)
SO SÁNH
\(\frac{2^{2017}+1}{2^{2014}+1}>1\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+\left(1+3\right)}{2^{2014}+\left(1+3\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+4}{2^{2014}+4}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{4\left(2^{2015}+1\right)}{4\left(2^{2012}+1\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2015}+1}{2^{2012}+1}\)
Đặt
\(A=\frac{2^{2015}+1}{2^{2012}+1}\) và \(B=\frac{2^{2017}+1}{2^{2014}+1}.\)
Ta có:
\(\frac{1}{8A}=2^{2015}+\frac{1}{2^{2015}}+8=2^{2015}+8-\frac{7}{2^{2015}}+8=1-\frac{7}{2^{2015}}+8.\)
\(\frac{1}{8B}=2^{2017}+\frac{1}{2^{2017}}+8=2^{2017}+8-\frac{7}{2^{2017}}+8=1-\frac{7}{2^{2017}}+8.\)
Vì \(2^{2015}< 2^{2017}.\)
\(\Rightarrow\frac{7}{2^{2015}}>\frac{7}{2^{2017}}.\)
\(\Rightarrow\frac{7}{2^{2015}}+8>\frac{7}{2^{2017}}+8.\)
\(\Rightarrow1-\frac{7}{2^{2015}}+8< 1-\frac{7}{2^{2017}}+8.\)
\(\Rightarrow A< B.\)
Hay \(\frac{2^{2015}+1}{2^{2012}+1}< \frac{2^{2017}+1}{2^{2014}+1}.\)
Chúc bạn học tốt!
So sánh (2^2015) + 1 / (2^2012) + 1 và (2^2017) + 1 / (2^2014) + 1
Có nhiều cách giải bài này. Hiện tôi có cách giải như sau tôi nghĩ là nó là ngắn nhất
Đặt: (2^2015)+1/(2^2012)+1 là A và (2^2017)+1/(2^2014)+1 là B
1/8A=(2^2015)+1/(2^2015)+8=(2^2015)+8-7/(2^2015)+8=1-7/(2^2015)+8
1/8B=(2^2017)+1/(2^2017)+8=(2^2017)+8-7/(2^2017)+8=1-7/(2^2017)+8
Vì 2^2015+8<2^2017+8 nên 7/(2^2015+8)>7/(2^2017)+8 nên 1-7/(2^2015)+8<1-7/(2^2017)+8 từ đó suy ra B>A hay 2^2017+1/(2^2014)+1>(2^2015)+1/(2^2012)+1
SO SÁNH:
A=\(\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2016}+\frac{1}{2017}}\)
VÀ
B=2017
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
So Sánh:
2^2015+1/2^2012+1 và 2^2017+1/2^2014+1
So sánh : \(\frac{2^{2015}+1}{2^{2012}+1}=\frac{2^{2017}+1}{2^{2014}+1}\)
Đặt \(A=\frac{2^{2015}+1}{2^{2012}+1}\) và \(B=\frac{2^{2017}+1}{2^{2014}+1}\)
Ta có: \(\frac{1}{8A}=2^{2015}+\frac{1}{2^{2015}}+8=2^{2015}+8-\frac{7}{2^{2015}}+8=1-\frac{7}{2^{2015}}+8\)
\(\frac{1}{8B}=2^{2017}+\frac{1}{2^{2017}}+8=2^{2017}+8-\frac{7}{2^{2017}}+8=1-\frac{7}{2^{2017}}+8\)
Ta có: \(7^{2015}< 7^{2017}\)
\(\Rightarrow\frac{7}{2^{2015}}>\frac{7}{2^{2017}}\)
\(\Rightarrow1-\frac{7}{2^{2015}}+8< 1-\frac{7}{2^{2017}}+8\)
hay A<B
hay \(\frac{2^{2015}+1}{2^{2012}+1}\)<\(\frac{2^{2017}+1}{2^{2014}+1}\)
So sánh:\(\frac{2^{2014}+1}{2^{2016}+1}và\frac{2^{2015+1}}{2^{2017}+1}\)
So Sánh :
\(\frac{2^{1015}+1}{2^{2012}+1}\)và \(\frac{2^{2017}+1}{2^{2014}+1}\)
Đặt: (2^2015)+1/(2^2012)+1 là A và (2^2017)+1/(2^2014)+1 là B
1/8A=(2^2015)+1/(2^2015)+8=(2^2015)+8-7/(2^2015)+8=1-7/(2^2015)+8
1/8B=(2^2017)+1/(2^2017)+8=(2^2017)+8-7/(2^2017)+8=1-7/(2^2017)+8
Vì 2^2015+8<2^2017+8 nên 7/(2^2015+8)>7/(2^2017)+8 nên 1-7/(2^2015)+8<1-7/(2^2017)+8 từ đó suy ra B>A hay 2^2017+1/(2^2014)+1>(2^2015)+1/(2^2012)+1
1.So sánh:
\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2014}\) và \(4\)
2. Tính :
\(\left(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\)
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!