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\(\dfrac{a}{2016}=\dfrac{b}{2017}=\dfrac{c}{2018}=\dfrac{a-c}{2016-2018}=\dfrac{a-b}{2016-2017}=\dfrac{b-c}{2017-2018}\)
\(\rightarrow\dfrac{a-c}{-2}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}\)
\(\rightarrow a-c=2\cdot\left(a-b\right)=2\cdot\left(b-c\right)\)
\(\rightarrow\left(a-c\right)^3=\left[2\cdot\left(a-b\right)\right]^2\cdot2\cdot\left(b-c\right)\)
\(\Rightarrow\left(a-c\right)^3=8\cdot\left(a-b\right)^2\cdot\left(b-c\right)\)
Ta có: \(P=\frac{6^{2017}.4^{2018}.75^{1009}}{2^{4035}.3^{3025}.10^{2018}}=\frac{\left(2.3\right)^{2017}.\left(2^2\right)^{2018}.\left(5.5.3\right)^{1009}}{2^{4035}.3^{3025}.\left(2.5\right)^{2018}}\)
\(=\frac{2^{2017}.3^{2017}.2^{4036}.5^{2018}.3^{1009}}{2^{4035}.3^{3025}.2^{2018}.5^{2018}}=\frac{2^{6053}.3^{3026}.5^{2018}}{2^{6053}.3^{3025}.5^{2018}}=3\)
Vậy P=3 <=> A. P=3
Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\Rightarrow a=2016k;b=2017k;c=2018k\)
\(\frac{a}{24}+\frac{b}{4}=\frac{c}{2018}\)
\(\Rightarrow\frac{2016k}{24}+\frac{2017k}{4}=\frac{2018k}{2018}\)
\(\Rightarrow84k+504,25k=k\)
\(\Rightarrow k=0\)
\(\Rightarrow a,b,c=0\)
Đặt a/2016 = b/2017 = c/2018 = k => a=2016k
b=2017k
c=2018k
Ta có (a-c)^3=( 2016k-2018k)^3 = (k(2016-2018))^3 = (k(-2))^3 (1)
Ta lại có 8(a-b)^2*(b-c)= 8(2016k-2017k)^2*(2017k-2018k) = 8(k(2016-2017)^2*(k(2017-2018) = 2^3*(k(-1))^2*(k(-1)) = 2^3*k^2*1*k*(-1) = k^3*(-2)^3 = (k(-2))^3 (2)
Từ (1) và (2) suy ra (a-c0^3 = 8(a-b)^2*(b-c)
Nhớ tick mik nha 
Đặt a/2016=b/2017=c/2018=k
=>a=2016k; b=2017k; c=2018k
(a-c)^3=(2016k-2018k)^3=(-2k)^3=-8k^3
8(a-b)^2*(a-b)
=8(a-b)^3
=8(2016k-2017k)^3
=-8k^3
=(a-c)^3
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=1\Leftrightarrow a=b\\\dfrac{b}{c}=1\Leftrightarrow b=c\\\dfrac{c}{a}=1\Leftrightarrow c=a\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow A=\dfrac{a^{2017}\cdot a^{2018}}{c^{4035}}=\dfrac{a^{2017}\cdot a^{2018}}{a^{4035}}=\dfrac{a^{4035}}{a^{4035}}=1\)