ta cs: \(\frac{a+2006}{a-2006}=\frac{b+2005}{b-2005}\)
\(\Rightarrow\frac{a+2006}{b+2005}=\frac{a-2006}{b-2005}=\frac{a}{b}=\frac{2006}{2005}\)
=> dpcm
Các câu hỏi tương tự
Cho frac{a}{b}frac{c}{d}. Chứng minh:a) frac{left(a-bright)^3}{left(c-dright)^3}frac{3a^2+2b^2}{3c^2+2d^2}b)frac{4a^4+5b^4}{4c^4+5d^4}frac{a^2b^2}{c^2d^2}c)left(frac{a-b}{c-d}right)^{2005}frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}d)frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}frac{left(a+bright)^{2005}}{left(c+dright)^{2005}}e)frac{left(20a^{2006}+11b^{2006}right)^{2007}}{left(20a^{2007}-11b^{2007}right)^{2006}}frac{left(20c^{2006}+11d^{2006}right)^{2007}}{left(20c^{2007}-11d^{2007}right)^{20...
Đọc tiếp
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
so sánh: \(\left(\frac{2006-2005}{2006+2005}\right)^2\) và \(\frac{2006^2-2005^2}{2006^2+2005^2}\)
a=\(\frac{2005^{2005}+1}{2005^{2006}+1}\) va b=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)
so sánh a và b
cho \(\frac{a+2006}{a-2006}\)=\(\frac{b+2005d}{b-2005d}\) Chứng minh \(\frac{a}{b}\)=\(\frac{2006}{2005}\)
C=\(\frac{\frac{2006}{2}}{\frac{2006}{1}}\) +\(\frac{2006}{\frac{3}{\frac{2005}{2}}}\) +\(\frac{2006}{\frac{4}{\frac{2004}{3}}}\) +...+\(\frac{2006}{\frac{2007}{\frac{1}{2006}}}\)
Cho a/b=c/d chứng tỏ (2005.a-2006.b)/(2006.c-2007.d)=(2005.c-2006.d)/(2006.a-2007.b)
so sánh A và B biết :
A = \(\frac{^{10^{2004+1}}}{10^{2005+1}}\)B= \(\frac{10^{2005+1}}{10^{2006+1}}\)
a)\(\frac{x-1}{2005}=\frac{3-y}{2006}\)và x-y=4009
b)\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\)và x-y-z=28
Chứng minh rằng :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}_{ }\)