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\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Ta có:
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Cộng vế theo vế, ta có:
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Vậy A > B
a, Bn quy đồng rồi làm nha
b,Có A=2017^2017+1/2017^2018+1
--> 2017A=2017^2018+2017/2017^2018+1
2017A=2017^2018+1/2017^2018+1 + 2016/2017^2018+1
2017A=1+ 2016/2017^2018+1
Có B=2017^2016+1/2017^2017+1
--> 2017B=2017^2017+2017/2017^2017+1
2017B=2017^2017+1/2017^2017+1 + 2016/2017^2017+1
2017B=1+2016/2017^2017+1
Vì 1+2016/2017^2018+1 < 1+2016/2017^2017+1
nên 2017A<2017B
-->A<B
A.Ta có :
\(A=-\frac{15}{46}>-\frac{15}{45}=-\frac{51}{153}>-\frac{51}{151}=B\)
\(\Rightarrow A>B\)
A=1+2917+20172+20173+.....+201748+201749
Đặt: C = 20172+ 20173+.....+201748+201749
=> 2017C =20173+.20174....+201749+201750
=> 2017C-C = (20173+.20174....+201749+201750 ) -(20172+ 20173+.....+201748+201749 )
=> 2016C = 201750- 20172 => C= (201750- 20172)/2016
=> A = 1+2917 + (201750- 20172)/2016 < 2017^50-1 = B
2A=2+20172+20173+20174+...+201749+201750
2A-A=201750-1
A=201750-1. Vậy A=B
Câu B
201750-1=20174.12+2-1=(20174)12.20172-1=A112.S9-1=B1.S9-1=X9-1=F8
Tính A và B rồi ta đi so sánh:
A = \(\frac{2016}{2017}\) + \(\frac{2017}{2018}\) = \(1.999008674\)
B = \(\frac{2016+2017}{2017+2018}\) = \(0.9995043371\)
Mà 1.999008674 > 0.9995043371
Nên: A > B
ta có: a+2017/b+2017=a/b
=>a/b=a/b
hay a/b=a+2017/b+2017
+Nếu a=b=>a/b=1(1)
Do a=b=>a+2017=b+2017=>a+2017/b+2017=1(2)
Từ (1)và(2)=>a/b=a+2017/b+2017
+Nếu a/b<1=>a<b=>2017a<2017b
=>ab+2017a<ab+2017b
=>a(b+2017)<b(a+2017)
=>a/b<a+2017/b+2017
+Nếu a/b>1=>a>b=>2017a>2017b
=>ab+2017a>ab+2017b
=>a(b+2017)>b(a+2017)
=>a/b>a+2017/b+2017
*
thanhs các bạn nha