Cho a,m,n \(\in\) N*, hãy so sánh hai tổng sau : A= \(\frac{10}{a^m}+\frac{10}{a^n}\) và B= \(\frac{11}{a^m}+\frac{9}{a^n}\)
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Cho a , m ,n thuộc N sao , Hãy So sánh :
\(A=\frac{10}{a^m}+\frac{10}{a^n}\&B=\frac{11}{a^m}+\frac{9}{a^n}\)
Cho a,m,n \(\in\)N* . Hãy so sánh A và B :
A = \(\frac{10}{a^m}+\frac{10}{a^n}\)
B = \(\frac{11}{a^m}+\frac{9}{a^n}\)
Cho a,m,n \(\in\)N* . Hãy so sánh :A=\(\frac{10}{a^m}+\frac{10}{a^n}\)và B=\(\frac{11}{a^m}+\frac{9}{a^n}\)
( Ai giải mình mới tick nha )
Cho a,m,n inN* , sosánhhaitổngsau:Afrac{10}{a^m}+frac{10}{a^n}và Bfrac{11}{a^m}+frac{9}{a^n}
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\(Cho\) \(a,m,n\) \(\in\)N* , \(so\)\(sánh\)\(hai\)\(tổng\)\(sau\)\(:\)
\(A=\frac{10}{a^m}+\frac{10}{a^n}\)\(và\) \(B=\frac{11}{a^m}+\frac{9}{a^n}\)
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\(B-A=\frac{11-10}{a^m}+\frac{9-10}{a^n}=\frac{1}{a^m}-\frac{1}{a^n}\)
Nếu \(m>n\) thì \(\frac{1}{a^m}-\frac{1}{a^n}< 0\Rightarrow B< A\)
Nếu \(m< n\) thì \(\frac{1}{a^m}-\frac{1}{a^n}>0\Rightarrow B>A\)
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so sánh \(A=\frac{10}{a^m}+\frac{10}{a^n};B=\frac{11}{a^m}+\frac{9}{a^n}\)
So sánh: A=\(\frac{10}{a^m}+\frac{10}{a^n}\) và \(B=\frac{11}{a^m}+\frac{9}{a^n}\)
A) So sánh :Cho a,m,nin N*,hãy so sánh A và B:afrac{10}{a^m}+frac{10}{a^n} bfrac{11}{a^m}+frac{9}{a^n} B) tìm x,biết frac{x-1}{2011}+frac{x-2}{2012}+frac{x-3}{2013}frac{x-4}{2014}+frac{x-5}{2015}+frac{x-6}{2016}
Đọc tiếp
A) So sánh :Cho a,m,n\(\in\) N*,hãy so sánh A và B:
\(a=\frac{10}{a^m}+\frac{10}{a^n}\) \(b=\frac{11}{a^m}+\frac{9}{a^n}\)
B) tìm x,biết \(\frac{x-1}{2011}+\frac{x-2}{2012}+\frac{x-3}{2013}=\frac{x-4}{2014}+\frac{x-5}{2015}+\frac{x-6}{2016}\)
B,
(1 - x-1/2011)+(1 - x-2/2012)+(1 - x-3/2013)=(1 - x-4/2014)+(1 - x-5/2015)+(1 - x-6/2016)
=> 2010-x/2011 + 2010-x/2012 + 2010-x/2013 = 2010-x/2014 + 2010-x/2015 + 2010-x/2016
=> 2010-x/2011 + 2010-x/2012 + 2010-x/2013 - 2010-x/2014 - 2010-x/2015 - 2010-x/2016=0
=>(2010-x).(1/2011 + 1/2012 + 1/2013 + 1/2014 - 1/2015 - 1/2016)=0
Mà: 1/2011 + 1/2012 + 1/2013 + 1/2014 - 1/2015 - 1/2016 khác 0
=> 2010-x=0
=>x=2010
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a, 10/a^m > 11/a^m; 10/a^n > 9/a^n => A > B
b, bạn cộng 1 vào các phân số đưa VP qua VT đặt nhân tử chung x + 2010 thì trong ngoặc còn lại là số dương nên x + 2010 = 0
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a, Cho a,b,n \(\in\)N* Hãy so sánh \(\frac{a+n}{b+n}\)và \(\frac{a}{b}\)
b, Cho \(A=\frac{10^{11}-1}{10^{12}-1};B=\frac{10^{10}+1}{10^{11}+1}\)So sánh A và B
a, Cho a,b,n\(\in\)N*. Hãy so sánh \(\frac{a+n}{b+n}và\frac{a}{b}\)
b, Cho \(A=\frac{10^{11}-1}{10^{12}-1}\); \(B=\frac{10^{10}+1}{10^{11}+1}\). So sánh A và B
b)A=10^11-1/10^12-1
=> A< (10^11-1)+11/(10^12-1)+11=10^11+10/10^12+10=10.(10^10+1)/10.(10^11+1)=10^10+1/10^11+1<B
Vậy A<B
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