Các câu hỏi tương tự
\(Cho\)\(abc\ne0,và\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right).\)
cho các số hữu tỉ a,b,c thỏa mãn\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)và\(a+b+c\ne0\)tính gt bt\(P=\left(1+\frac{2a}{b}\right)\left(1+\frac{2b}{c}\right)\left(1+\frac{2c}{a}\right)\)
Cho \(abc\ne0\) và \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
Cho \(\frac{a}{b}\)= \(\frac{b}{c}\)= \(\frac{c}{d}\)= \(\frac{d}{a}\). Tính M = \(\frac{\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)}{a.b.c.d}\)
31 tháng 10 2020 lúc 22:16
Cho các số \(a,b,c\ne0\) thỏa mãn: \(\frac{ab}{a+b}=\frac{bc}{b+c}+\frac{ac}{a+c}\).
Tính \(\left(1+\frac{a}{2b}\right)\left(1+\frac{b}{3c}\right)\left(1+\frac{c}{4a}\right)\)
Cho \(abc\ne0\) và \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
cho a,b,c thỏa mãn : \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-b\right)\left(c-a\right)}=2013\)
tính M = \(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)
cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\left(a,b,c\ne0;b\ne c\right)\)) chứng minh rằng : \(\frac{a}{b}=\frac{a-c}{c-b}\)
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\left(a,b,c\ne0,b\ne c\right)\).Chứng minh rằng\(\frac{a}{b}=\frac{a-c}{c-b}\)