Tính A+B
A= 1.99+2.98+3.97+.............+99.1
B=1.101+2.102+3.103+............+99.199
so sánh P và Q
P=2016/2017+2017/2018
Q= 2016+2017+2018/2017+2018+2019
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A+B = (1.99+2.98+3.97+...+99.1)+(1.101+2.102+3.103+...+99.199)
A+B = (1.99+1.101)+(2.98+2.102)+(3.97+3.103)+...+(99.1+99.199)
A+B = 1(99+101) + 2(98+102) + 3(97.103)+...+99(1+199)
A+B = 1.200 + 2.200 + 3.200 +...+ 99.200
A+B = 200.(1+2+3+...+200)
A+B = 200.4950
A+B = 990000
A + B = ( 1 . 99 + 2 . 98 + 3 . 97 + ... + 99 . 1 ) + ( 1 . 101 + 2 . 102 + 3 . 103 + ... + 99 . 199 )
A + B = 99 . ( 1 + 199 ) + 98 . ( 2 + 198 ) + 97 . ( 3 + 197 ) + ... + 2 . ( 102 + 98 ) + 1 . ( 99 + 101 )
A + B = 99 . 200 + 98 . 200 + 97 . 200 + ... + 2 . 200 + 1 . 200
A + B = ( 99 + 98 + 97 + ... + 2 + 1 ) . 200
A + B = 4950 . 200
A + B = 990000
Ta có :
\(\frac{2016}{2017}>\frac{2016}{2017+2018+2019}\)
\(\frac{2017}{2018}>\frac{2017}{2017+2018+2019}\)
\(\frac{2018}{2019}>\frac{2018}{2017+2018+2019}\)
\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}>\) \(\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)
\(\Rightarrow P>\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow P>Q\)
Chúc bạn học tốt !!!
vì P có các số bé hơn 1 còn Q có các số lớn hơn 1 =>P<Q
Vậy P<Q.
mình làm hơi tắt xin bạn thông cảm bạn tự viết các số có trong P;Q ra nhá
#)Giải :
\(Q=2+\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)
Ta thấy : \(2>\frac{2016}{2017};2>\frac{2017}{2018};2>\frac{2018}{2019}\left(1\right)\)
\(\frac{2016}{2017+2018+2019}< \frac{2016}{2017}\left(2\right)\)
\(\frac{2017}{2017+2018+2019}< \frac{2017}{2018}\left(3\right)\)
\(\frac{2018}{2017+2018+2019}< \frac{2018}{2019}\left(4\right)\)
Từ (1) (2) (3) (4) \(\Rightarrow P>Q\)
\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)
\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)
\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)
\(\Rightarrow A\)>\(3-1=2\)
\(B=\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow B=1-\frac{3}{6054}\)
\(\Rightarrow B=1-\frac{1}{2018}\)
\(B\)<\(1\);\(A\)>\(2\)
\(\Rightarrow A\)>\(B\)
A+B=(1.99+2.98+...+99.1)+(1.101+2.102+...+99.199)
=(1.99+1.101)+(2.98+2.102)+...+(99.1+99.199)
=1.(99+101)+2.(98+102)+...+99(1+199)
=200+2.200+...+99.200
=200.(1+2+3+4+...+99)
=200.4950
=.....