So sánh :
a,\(\frac{7}{23}v\text{à}\frac{11}{28}\)
b,\(\frac{2014}{2015}+\frac{2015}{2016}v\text{à}\frac{2014+2015}{2015+2016}\)
c,A=\(\frac{2^{10}+1}{2^{11}+1}v\text{à B=\frac{2^{11}+1}{2^{12}+1}}\)
so sánh a)\(\frac{10^{2014}-1}{10^{2015}-1}v\text{à}\frac{10^{2013}-1}{10^{2014}-1}\)
b) \(\frac{n+3}{n-2}v\text{à}\frac{n+5}{n-4}\)
1) CMR : A=(n+2015)(n+2016) + n2 + n chia hết cho 2 với n ϵ N
2) So sánh :
P = \(\frac{2013}{2014^{2013}}+\frac{2014}{2015^{2014}}+\frac{2015}{2016^{2015}}+\frac{2016}{2017^{2016}}\) và
Q = \(\frac{2014}{2017^{2016}}+\frac{2013}{2016^{2015}}+\frac{2016}{2015^{2014}}+\frac{2015}{2014^{2013}}\)
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)
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 A và B
a) A = \(\frac{10^{11}-1}{10^{12}-1}\) Và \(B=\frac{10^{10}+1}{10^{11}+1}\)
b) \(A=\frac{2000^{2015}+1}{2000^{2016}+1}\) Và \(B=\frac{2000^{2014}+1}{2000^{2015}+1}\)
b, 2000A = \(\frac{2000\left(2000^{2015}+1\right)}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+2000}{2000^{2016}+1}\)
= \(\frac{\left(2000^{2016}+1\right)+1999}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+1}{2000^{2016}+1}\) + \(\frac{1999}{2000^{2016}+1}\)
= 1 + \(\frac{1999}{2000^{2016}+1}\)
2000B = \(\frac{2000\left(2000^{2014}+1\right)}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+2000}{2000^{2015}+1}\)
= \(\frac{\left(2000^{2015}+1\right)+1999}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+1}{2000^{2015}+1}\) + \(\frac{1999}{2000^{2015}+1}\)
= 1 + \(\frac{1999}{2000^{2015}+1}\)
So sanh
câu b tiếp
So sánh 2000A với 2000B
Vì \(\frac{1999}{2000^{2016}+1}\) < \(\frac{1999}{2000^{2015}+1}\)
→ 2000A< 2000B
→ A<B
So sánh mà không cần tính \(\frac{2015}{1000}v\text{à}\frac{3030}{2015}\)
Ta có:
\(\frac{2015}{1000}=\frac{1000}{1000}+\frac{1015}{1000}=1+\frac{1015}{1000}\)
\(\frac{3030}{2015}=\frac{2015}{2015}+\frac{1015}{2015}=1+\frac{1015}{2015}\)
Vì \(\frac{1015}{1000}>\frac{1015}{2015}\)
=> \(\frac{2015}{1000}>\frac{3030}{2015}\)
Ủng hộ mk nha ^_-
\(\frac{2015}{1000}=2+\frac{15}{1000}\)
\(\frac{3030}{2015}=1+\frac{1015}{2015}\)
suy ra \(\frac{2015}{1000}>\frac{3030}{2015}\)
\(\frac{2015}{1000}>\frac{3030}{2015}\)
a, Cho A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\) . So Sánh A với 1
b, B=\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\). So sánh B với \(\frac{1}{2}\)
c, cho M=\(\frac{2013}{2014}+\frac{2014}{2015}\)và N=\(\frac{2013+2014}{2014+2015}\). So sánh M và N
Câu a, p/s cuối cùng là \(\frac{1}{100}\)nha mí bn
a) Ta có :
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)
\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)
vậy A > 1
b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)
\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)
Vậy B > \(\frac{1}{2}\)
So sánh : \(A=\frac{2015^{2016}+1}{2015^{2015}+1}\) và \(B=\frac{2014^{2015}+1}{2014^{2014}+1}\)
A = \(\frac{2015^{2016}+1}{2015^{2015}+1}=\frac{2015^{2015}+1}{2015^{2015}+1}+\frac{2015}{2015^{2015}+1}=1+\frac{2015}{2015^{2015}+1}\)
B = \(\frac{2014^{2015}+1}{2014^{2014}+1}=\frac{2014^{2014}+1}{2014^{2014}+1}+\frac{2014}{2014^{2014}+1}=1+\frac{2014}{2014^{2014}+1}\)
Rồi bạn tự so sánh nha
Bài 1:Một xe tải chạy từ A và có thể đến B sau 6 giờ .Sau khi xe tải chạy được 2 giờ thì 1 xe ô tô con khởi hành từ B chạy về A và gặp xe tải sau 1 giờ 36 phút .tính thời gian xe con chạy từ B về A
Bài 2:Tìm x biết
a,\(\frac{x}{6}\)\(+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}=\frac{220}{39}\)
b,\(2+4+6+....+2x=156\)
Bài 3:So sánh
a,A=\(\frac{2015}{x^m}+\frac{2015}{x^n}\)Và B=\(\frac{2014}{x^m}+\frac{2016}{x^n}\)
b,C=\(\frac{2015\cdot2016-1}{2015\cdot2016}\)Và D=\(\frac{2016\cdot2017-1}{2016\cdot2017}\)
c,E=\(\frac{5}{10^{2015}}+\frac{11}{10^{2016}}\)Và F=\(\frac{11}{10^{2015}}+\frac{5}{10^{2016}}\)
d,S=\(\frac{1}{2}+\frac{2}{2^3}+.....+\frac{n}{2^n}+.....+\frac{2017}{2^{2017}}\) \(với\)\(2\)
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!