\(\)\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)

\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)

\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)

\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)

\(\Rightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)

Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)

Nên:

\(x+2016=0\Rightarrow x=-2016\)

\(\left(4x-3\right)^2=\left(4x-3\right)\)

\(\Rightarrow\left(4x-3\right)^2-\left(4x-3\right)=0\)

\(\Rightarrow\left(4x-3\right)\left[\left(4x-3\right)-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\4x-3-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=3\\4x=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=1\end{matrix}\right.\)

Gọi số cần tìm là \(a\)

Theo đề bài ta có:

\(a:36\) dư \(24\)

Nên:

\(a=36k+24\)

\(\left\{{}\begin{matrix}36k⋮2\\24⋮2\end{matrix}\right.\)

Nên \(a⋮2\)

Ta có đpcm

Lớn nhất nhé bạn :D

\(A=\dfrac{1}{\left|x-y\right|+\left|y-z\right|+\left|z-x\right|+2016}\)

\(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y-z\right|\ge0\\\left|z-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-y\right|+\left|y-z\right|+\left|z-x\right|\ge0\)

\(\Rightarrow\left|x-y\right|+\left|y-z\right|+\left|z-x\right|+2016\ge2016\)

\(A=\dfrac{1}{\left|x-y\right|+\left|y-z\right|+\left|z-x\right|+2016}\le\dfrac{1}{2016}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y-z\right|=0\\\left|z-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)

\(\Rightarrow x=y=z\)

\(\left|x-3\right|+\left|x-5\right|=\left|x-3\right|+\left|5-x\right|\)

Áp dụng bđt:

\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\left|x-3\right|+\left|5-x\right|\ge\left|x-3+5-x\right|=2\)
Dấu "=" xảy ra khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\Rightarrow x\ge3\\5-x\ge0\Rightarrow x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\Rightarrow x< 3\\5-x< 0\Rightarrow x>5\end{matrix}\right.\end{matrix}\right.\)

Vậy \(3\le x\le5\)

\(A=-\left|x\right|\le0\)

Dấu "=" xảy ra khi:

\(x=0\)

\(B=4-\left|5x-2\right|\le4\)

Dấu "=" xảy ra khi:

\(\left|5x-2\right|=0\Rightarrow x=\dfrac{2}{5}\)

\(C=\left|x-3\right|-\left|5-x\right|=\left|x-3\right|-\left|x-5\right|\)

\(C\le\left|x-3-x-5\right|=8\)

Dấu "=" xảy ra khi:

\(3\le x\le5\)

\(D=4-\left|5x-2\right|-\left|3y+12\right|\le4\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|5x-2\right|=0\\\left|3y+12\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

\(E=5-3\left(2x-1\right)\le5\)

Dấu "=" xảy ra khi:

\(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

\(\left\{{}\begin{matrix}2^{101}>2^{100}=\left(2^{10}\right)^{10}=1024^{10}\\5^{39}< 5^{40}=\left(5^4\right)^{10}=625^{10}\end{matrix}\right.\)

\(1024^{10}>625^{10}\Rightarrow2^{101}>5^{39}\)

\(\left(2x+3\right)^2=\dfrac{9}{121}\)

\(\Rightarrow\left(2x+3\right)^2=\left(\pm\dfrac{3}{11}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=\dfrac{3}{11}\\2x+3=-\dfrac{3}{11}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{30}{11}\\2x=-\dfrac{36}{11}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{11}\\x=-\dfrac{18}{11}\end{matrix}\right.\)

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Leftrightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-20}{-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.4=20\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{3}{4}\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\)

\(\Rightarrow\dfrac{-3x}{-9}=\dfrac{5y}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{-3x}{-9}=\dfrac{5y}{20}=\dfrac{-3x+5y}{-9+20}=\dfrac{33}{11}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.3=9\\y=3.4=12\end{matrix}\right.\)