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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!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
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!
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
Ta có 2A= 2(2^2015 + 1)/ 2^2016 + 1 = 2^2016 +2 / 2^2016 +1 = 2^2016+1/2^2016+1 + 1/2^2016 +1= 1 + 1/2^2016
2B= 2( 2^2016 + 1/ 2^2017+ 1) = 2^2017 +2 / 2^2017 +1 = 2^2017+1/2^2017+1 + 1/2^2017 +1 = 1 + 1/2^2017
Do 1/2^2016 > 1/2^2017 => 2A>2B => A>B
10.A=\(10.A=\frac{10.\left(2^{2015+1}\right)}{2^{2016}+1}=\frac{2^{2016+10}}{2^{2016}+1}=1+\frac{2016}{2^{2016}+1}\)
\(10.B=\frac{10.\left(2^{2016}+1\right)}{2^{2017}+1}=\frac{2^{2017}+10}{2^{2017}+1}=1+\frac{2016}{2^{2017}+1}\)
Ta có:\(\frac{2016}{2^{2016}+1}>\frac{2016}{2^{2017}+1}\)
A = (n + 2015)(n + 2016) + n2 + n
= (n + 2015)(n + 2015 + 1) + n(n + 1)
Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2
=> (n + 2015)(n + 2015 + 1) chia hết cho 2
n(n + 1) chia hết cho 2
=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2
=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)
Đặt C = 1 + 2017 + 20172 + ... + 20172016 ; D = 1 + 2016 + 20162 + ... + 20162016
Ta có : 2017C = 2017 + 20172 + 20173 + ... + 20172017
=> 2016C = 2017C - C = 20172017 - 1\(\Rightarrow C=\frac{2017^{2017}-1}{2016}\)
2016D = 2016 + 20162 + 20163 + ... + 20162017
=> 2015D = 2016D - D = 20162017 - 1\(\Rightarrow D=\frac{2016^{2017}-1}{2015}\)
\(\Rightarrow A=\frac{2017^{2017}}{\frac{2017^{2017}-1}{2016}}=\frac{2017^{2017}.2016}{2017^{2017}-1}\);\(B=\frac{2016^{2017}}{\frac{2016^{2017}-1}{2015}}=\frac{2016^{2017}.2015}{2016^{2017}-1}\)
Ta có : 20172017.2016.(20162017 - 1) - 20162017.2015.(20172017 - 1)
= 20172017.20162017.2016 - 20172017.2016 - 20172017.20162017.2015 + 20162017.2015
= 20172017.20162017 - 20172017.2016 + 20162017.2015
= 20172017.(20162017 - 2016) + 20162017.2015 > 0
=> A > B
Ta có
\(A=1:\frac{1+2017+2017^2+...+2017^{2016}}{2017^{2017}}\)
\(B=1:\frac{1+2016+2016^2+...2016^{2016}}{2016^{2017}}\)
\(A=1:\left(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}\right)\)
\(B=1:\left(\frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\right)\)
Có 20172017>20162017 ; 20172016>20162016 ; 20172015>20162015;..... ; 2017>2016
=> \(\frac{1}{2017^{2017}}< \frac{1}{2016^{2017}};\frac{1}{2017^{2016}}< \frac{1}{2016^{2016}};\frac{1}{2017^{2015}}< \frac{1}{2016^{2015}};...;\frac{1}{2017}< \frac{1}{2016}\)
=> \(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}< \frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\)
=> A>B ( vì số bị chia và số chia của A và B đều dương, số bị chia của cả 2 đều là 1, cái nào có số chia nhỏ hơn thì lớn hơn)
