Question

[Nguyễn Hải Yến]

Bài 1: 

a)  Cho a(y+z) = b(z+c) = c(x+y)  Tính: \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-c}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)

b) \(Cho\dfrac{a}{2014}=\dfrac{b}{2015}=\dfrac{c}{2016}cm:4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

c) \(\dfrac{a}{a'}+\dfrac{b'}{b}=1\) và \(\dfrac{b}{b'}+\dfrac{c'}{c}=1\)

cm: abc+a'b'c'=0

bài 4: 

a) \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)    Tính: \(\dfrac{x}{y}\)

b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)  Tính P = \(\dfrac{xy+yz+xz}{x^2+y^2-z^2}\)

c) \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)

Tính : P = \(\dfrac{a+b}{c+d}+\dfrac{c+b}{a+d}=\dfrac{c+d}{a+b}=\dfrac{a+d}{c+b}\)

d) \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\) Tính: \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)