Gọi hai số \(\ne0\) cần tìm là \(x,y\)

Ta có: \(\dfrac{x+y}{3}=\dfrac{x-y}{\dfrac{1}{3}}=\dfrac{xy}{\dfrac{200}{3}}=\dfrac{x+y+x-y}{3+\dfrac{1}{3}}=\dfrac{2x}{\dfrac{10}{3}}=\dfrac{3x}{5}=k\ne0\)

\(\Rightarrow x=\dfrac{5k}{3}\left(1\right);x+y=3k\left(2\right)lxy=\dfrac{200k}{3}\left(3\right)\)

Từ (1) và (2) \(\Rightarrow y=3k-\dfrac{5k}{3}=\dfrac{4k}{3}\Rightarrow xy=\dfrac{5k}{3}.\dfrac{4k}{3}=\dfrac{20k^2}{9}\left(4\right)\)

Từ (3) và (4) \(\Rightarrow\dfrac{200k}{3}=\dfrac{20k^2}{9}\Leftrightarrow k=30\left(k\ne0\right)\Rightarrow x=\dfrac{5.30}{3}=50;y=\dfrac{4.30}{3}=40\)

Vậy hai số cần tìm là 40; 50

Từ \(a^2+a-p=0\Rightarrow p=a^2+a=a\left(a+1\right)\)

Với \(a\in Z\Rightarrow p=a\left(a+1\right)⋮2,p\) là số nguyên tố \(\Rightarrow p=2\)

\(\Rightarrow a\left(a+1\right)=2=1.2=\left(-1\right).\left(-2\right)\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\end{matrix}\right.\)

Từ \(b^2=ac\) và \(c^2=bd\) ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)

Mà \(\dfrac{a^3}{b^3}=\dfrac{a.a.a}{b.b.b}=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)

Từ (1) và (2) ta có đpcm 

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)

\(=\dfrac{4.4^5}{3.3^5}.\dfrac{6.6^5}{2.2^5}=\dfrac{4^6}{3^6}.\dfrac{6^6}{2^6}=\left(\dfrac{6}{3}\right)^6.\left(\dfrac{4}{2}\right)^6=2^{12}\)

\(\Rightarrow2^x=2^{12}\Rightarrow x=12\)

Nhận xét \(\left\{{}\begin{matrix}|2x-2011|\ge0\forall x\\\left(3y+2012\right)^{2012}\ge0\forall y\end{matrix}\right.\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}2x-2011=0\\3y+2012=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2011}{2}\\y=-\dfrac{2012}{3}\end{matrix}\right.\)

\(a,A=\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}=\dfrac{5.2^{2.15}.3^{2.9}-2^2.3^{20}.2^{3.9}}{5.2^{10}.2^{19}.3^{19}-7.2^{29}.3^{3.6}}\)

\(=\dfrac{2^{29}.3^{18}.\left(5.2-3^2\right)}{2^{29}.3^{18}.\left(5.3-7\right)}=\dfrac{10-9}{15-7}=\dfrac{1}{8}\)

\(b,B=1+49.\dfrac{1}{49}.\left(2^6:2^5\right)=1+2=3\)

Ta có \(A=2x^2-6x-x^2+7x-5x+2015\)

             \(=x^2-4x+2015\)

A, Với \(x=4\) ta được \(A=2015\)

B, \(A=2015\Rightarrow x^2-4x=0\Rightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)