\(\frac{5-a}{2010}+\frac{4-a}{2011}+\frac{3-a}{2012}=\frac{6-a}{2009}+\frac{1-a}{1007}\)
Tìm a
Những câu hỏi liên quan
Tìm a , biết :
\(\frac{5-a}{2010}+\frac{4-a}{2011}+\frac{3-a}{2012}=\frac{6-a}{2009}+\frac{1-a}{1007}\)
Không sử dụng máy tính, hãy so sánh A và B:
A=\(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012},B=\frac{1}{2009}+\frac{1}{1007}\)
VÌ A = 1/2010 > 1/2011 > 1/2012 (1)
B = 1/2009 <1/1007 (2)
TỪ (1) VÀ (2) => 1/2010 < 1/1007
VẬY A < B
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A có chia hết cho 3 không?
\(A=\frac{2012+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}}\)
Xét tử:
\(2012+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{1}{2012}\)
= \(\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1\)
= \(\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
= \(2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
Thay vào ta có:
A = \(\frac{2013\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}\)
=> A = 2013
Mà 2013 chia hết cho 3
=> A chia hết cho 3
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\(A=\frac{2012+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}}\)
hỏi A có chia hết cho 3 hay ko ?
http://d.f24.photo.zdn.vn/upload/original/2016/02/14/10/03/3204324726_616688374_574_574.jpg
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So sánh A và B biết
A=\(\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}\)
B=\(\frac{2009+2010+2011}{2010+2011+2012}\)
A=2.998508205
B=0.999502735
suy ra A>B
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Bài giải
Theo bài ra :
\(A=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}\)
\(B=\frac{2009+2010+2011}{2010+2011+2012}=\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
Ta có :
\(\frac{2009}{2010}>\frac{2009}{2010+2011+2012}\)
\(\frac{2010}{2011}>\frac{2010}{2010+2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}>\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }A>B\)
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Bài giải
Theo bài ra :
\(A=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}\)
\(B=\frac{2009+2010+2011}{2010+2011+2012}=\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
Ta có :
\(\frac{2009}{2010}>\frac{2009}{2010+2011+2012}\)
\(\frac{2010}{2011}>\frac{2010}{2010+2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}>\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }A>B\)
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Xem thêm câu trả lời
Tìm số hữu tỉ x biết:
a) \(\frac{x+4}{2009}+\frac{x+3}{2010}=\frac{x+2}{2011}+\frac{x+1}{2012}\)
b) \(\frac{x-2011}{2010}+\frac{x-2011}{2011}+\frac{x-2011}{2012}=\frac{x-2011}{2013}+\frac{x-2011}{2014}\)
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
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\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{2011}+\frac{1}{2012}\)
\(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\cdot\cdot\cdot\frac{2}{2010}+\frac{1}{2011}\)
Tính \(\frac{B}{A}\)
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
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từ dề bài ta có :
\(\frac{B}{A}=\frac{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=2012\)
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Tính
a)\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....\frac{1}{2011}}\)
b)\(\frac{0.375-0.3+\frac{3}{11}+\frac{3}{12}}{-0.265+0.5-\frac{5}{11}-\frac{5}{12}}+\frac{1.5+1-0.75}{2.5+\frac{5}{3}-1.25}\)
Tìm x biết
a)\(\frac{x+4}{2009}+\frac{x+3}{2010}=\frac{x+2}{2011}+\frac{x+1}{2012}\)
b)\(\left(\frac{1}{4}x-1\right)\)+\(\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+5\right)=3,5\)
Anh chỉ giải câu a thôi, câu b anh thấy nó bình thường mà.
Cộng vào mỗi phân số thêm 1 đơn vị được:
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\).
Tới đây tự làm tiếp nhá.
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