Các câu hỏi tương tự
Cho các số a, b, c khác 0 thỏa mãn b2= ac. Chứng minh rằng;
\(\frac{a}{c}=\frac{\left(2010a+2011b\right)^2}{\left(2010b+2011c\right)^2}\frac{ }{ }\)
Cho:\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)tinh:\frac{2011a-2010b}{c+d}=\frac{2011b-2010a}{c+d}=\frac{2011c+2011d}{a+b}=\frac{2011d-2010a}{c+dc=d}\)
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)\\ \frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}=?\\ aigiảigiúpmìnhvới\\ \)
Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)\)
Tính \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}=\frac{2011d-2010a}{b+c}\)
cho các số a,b,c khác 0 thỏa mãn b2=a*c.Cmr :
\(\frac{a}{c}=\frac{\left(2010a+2011b\right)^2}{\left(2010b+2011c\right)^2}\)Trình bày rõ ra nha. Ai làm được tớ tick cho
\(Cho\frac{a}{2b}=\frac{b}{2c}=\frac{d}{2a}\left(a,b,c>0\right)\)
\(TínhA=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)
cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a;b;c;d\ne0\right)\)
\(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}=?\)
tinhs A
cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a;b;c;d\ne0\right)\)
\(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}=?\)
tinhs A
cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a;b;c;d\ne0\right)\)
\(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}=?\)
tinhs A