Các câu hỏi tương tự
Cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\). CMR: 4(a-b)(b-c) = (c-a)\(^2\)
Cho \(\frac{a}{b}=\frac{c}{d}\)
CMR:\(\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)
CMR:\(\frac{a^{2005}}{b^{2005}}=\frac{\left(a-c\right)^{2005}}{\left(b-d\right)^{2005}}\)
Giúp với ạ(mn đừng giải bằng cách đặt k nha)
23 tháng 8 2017 lúc 20:27
cho :
\(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)
Chứng minh rằng : 4 . ( a - b ) . ( b - c ) = ( c - a )2
Cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\). Chứng minh rằng 4 .( a - b ) .( b - c ) = ( c - a )2
cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\) chứng minh rằng : \(4\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)
Cho \(\frac{a}{2003}\)=\(\frac{b}{2004}=\frac{c}{2005}\). Chứng minh rằng :\(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
\(cho:\frac{a^2+2004^2}{b^2+2005^2}=\frac{2004a}{2005b}\left(a,bkhac0\right).CMR:\orbr{\begin{cases}\frac{a}{2004}=\frac{b}{2005}\\\frac{a}{2004}=\frac{2005}{b}\end{cases}}\)
1/ Cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)
Chứng minh: \(4(a-b)\left(b-c\right)=\left(c-a\right)^2\)
Cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)
Chứng minh rằng: \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)