Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó : \(\frac{ac}{bd}=\frac{b.d.k^2}{b.d}=k^2\left(1\right);\)
\(\frac{2010a^2+2011c^2}{2010b^2+2011d^2}=\frac{2010b^2.k^2+2011d^2.k^2}{2010b^2+2011d^2}=\frac{k^2.\left(2010b^2+2011d^2\right)}{2010b^2+2011d^2}=k^2\left(2\right)\)
Từ (1)(2) => \(\frac{ac}{bd}=\frac{2010a^2+2011c^2}{2010b^2+2001d^2}\left(\text{đpcm}\right)\)
Cho:\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)tinh:\frac{2011a-2010b}{c+d}=\frac{2011b-2010a}{c+d}=\frac{2011c+2011d}{a+b}=\frac{2011d-2010a}{c+dc=d}\)
Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)\)
Tính \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}=\frac{2011d-2010a}{b+c}\)
a) Tìm ba số x, y z thỏa mãn : \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)và \(2x^2+2y^2-3z^2=-100\)
b) 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= ac. Chứng minh rằng;
\(\frac{a}{c}=\frac{\left(2010a+2011b\right)^2}{\left(2010b+2011c\right)^2}\frac{ }{ }\)
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)\\ \frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}=?\\ aigiảigiúpmìnhvới\\ \)
cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a;b;c;d\ne0\right)\)
\(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}=?\)
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
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
