so sán a và b biết a=\(\frac{2017\times2018-1}{2017\times2018}\)và b =\(\frac{2019\times2020-1}{2019\times2020}\)
So sánh: a)\(\frac{-3}{100}\)và \(\frac{2}{3}\) b)\(\frac{267}{-268}\)và\(\frac{-1347}{1343}\) c) \(\frac{2017\times2018-1}{2017\times2018}\)và\(\frac{2018\times2019-1}{2018\times2019}\) e)\(\frac{2017\times2018}{2017\times2018+1}\)và\(\frac{2018\times2019}{2018\times2019+1}\) Gải cách rút gọn cám ơn ạ
a) \(\frac{-3}{100}\)và\(\frac{2}{3}\) b)\(\frac{267}{-268}\)và\(\frac{-1347}{1343}\)c)\(\frac{2017\times2018-1}{2017\times2018}\)và\(\frac{2018\times2019-1}{2018\times2019}\)d)\(\frac{2017\times2018}{2017\times2018+1}\)và\(\frac{2018\times2019}{2018\times2019+1}\)
a) Ta có : \(\frac{-3}{100}< 0< \frac{2}{3}\)
\(\Rightarrow\frac{-3}{100}< \frac{2}{3}\)
b) Ta có : \(\frac{267}{268}< 1< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{268}< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{-268}< \frac{-1347}{1343}\)
c) Ta có : \(\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(\frac{2018.2019-1}{2018.2019}=\frac{2018.2019}{2018.2019}-\frac{1}{2018.2019}=1-\frac{1}{2018.2019}\)
mà \(2017.2018< 2018.2019\)
\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)
\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)
\(\Rightarrow\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
d) Ta có : \(\frac{2017.2018}{2017.2018+1}=\frac{2017.2018+1}{2017.2018+1}-\frac{1}{2017.2018+1}=1-\frac{1}{2017.2018+1}\)
\(\frac{2018.2019}{2018.2019+1}=\frac{2018.2019+1}{2018.2019+1}-\frac{1}{2018.2019+1}=1-\frac{1}{2018.2019+1}\)
mà \(2017.2018+1< 2018.2019+1\)
\(\Rightarrow\frac{1}{2017.2018+1}>\frac{1}{2018.2019+1}\)
\(\Rightarrow1-\frac{1}{2017.2018+1}< 1-\frac{1}{2018.2019+1}\)
\(\Rightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}\)
Tính gía trị của biểu thức
\(\frac{\left(2015^2\times2025+31\times2016-1\right)\times\left(2015\times2020+4\right)}{2016^2\times2017\times2018\times2019\times2020}\)
Tính nhanh:
a)\(\frac{131313}{151515}+ \frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)
b)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)
c)\(\frac{1}{3\times5}+\frac{1}{4\times7}+\frac{1}{7\times9}+...\frac{1}{2015\times2017}+\frac{1}{2017\times2018}\)
\(a,\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)
\(=\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=13\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{7.9}\right)\)
\(=13\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(=13.\frac{2}{9}=\frac{26}{9}\)
\(b,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
P/s :Dấu chấm là dấu nhân nha
\(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)
\(=\frac{13.10101}{15.10101}+\frac{13.10101}{35.10101}+\frac{13.10101}{63.10101}+\frac{13.10101}{99.10101}\)
\(=13.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=\frac{13}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}.\left(\frac{11}{33}-\frac{3}{33}\right)\)
\(=\frac{13}{2}.\frac{8}{33}\)
\(=\frac{52}{33}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
Sửa đề chút:
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2018}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)
B tự làm nốt nhé
Bài 1:So sánh:
\(A=2019\times2020\)và \(B=2020^2\times2019^2\)
DỄ THÔI BN Ạ
B =2020 . 2020 . 2019 . 2019
SUY RA A <B
1.Tính nhanh tổng sau:
\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)
2. So sánh A và B, biết:
\(A=\frac{2016^{2017}}{2016^{2017}-3}\)
\(B=\frac{2017^{2019}+1}{2017^{2019}-1}\)
1. Bài giải:
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\)
\(\Rightarrow\frac{1}{2}A=A-\frac{1}{2}A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1000}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\right)\)
\(\Rightarrow\frac{1}{2}A=1-\frac{1}{1002}=\frac{1001}{1002}\Rightarrow A=\frac{2002}{1002}=\frac{1001}{501}\)
Vậy \(A=\frac{1001}{501}\)
A = \(\frac{1}{1\times4}\)+ \(\frac{1}{4\times7}\)+ . . . . + \(\frac{1}{2017\times2020}\)
\(3A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{2017\cdot2020}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2020}\)
\(3A=1-\frac{1}{2020}\)
\(A=\frac{673}{2020}\)
Dấu '.' là dấu nhân nha
Học tốt~
Cho a(b+c)2+b(c+a)2+c(a+b)2=4abc và a2017+b2017+c2017=1
Tính A=\(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}\)
Ta có a(b+c)^2 +b(c+a)^2+c(a+b)^2 =4abc
ab^2+ac^2+2abc+ba^2bc^2+2abc+ca^2+cb^2+2abc=4abc
ab^2+ac^2+bc^2+ba^2+cb^2+ca^2+2abc=0
(ab^2+abc)+(ac^2+abc)+(bc^2+cb^2)+(a^2b+a^2c)=0
ab(b+c)+ac(b+c)+bc(b+c)+a^2(b+c)=0
(b+c)(ab+ac+bc+a^2)=0
(b+c)(a+b)(a+c)=0
*th1:b+c=0=> b=-c
=> b^2017 +c^2017 =0
mà a^2017 +b^2017 +c^2017=1
=>a^2017=1 => a=1
thay vào A rồi dc A=1
các th khác tương tự
So Sánh
A=\(\frac{11}{2017}+\frac{4}{2019}và\)B=\(\frac{10}{2017}+\frac{10}{2019}\)
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}\Rightarrow2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)
\(B=\frac{2016^{2017}-3}{2016^{2018}-3}\Rightarrow2016B=\frac{2016^{2018}-6048}{2016^{2018}-3}=1+\frac{-6045}{2016^{2018}-3}\)
Vì \(\frac{2015}{2016^{2017}+1}>0;\frac{-6045}{2016^{2018}-3}< 0\)
Nên: A>B