\(\left(3^x+1\right)+4.3^3=567\)

\(\left(3^x+1\right)+108=567\)

\(3^x+1=567-108\)

\(3^x+1=459\)

\(3^x=459-1\)

\(3^x=458\)

\(\Rightarrow\)Ko có giá trị của x

Ta có : \(\hept{\begin{cases}\left(x-3\right)^2\ge0\forall x\\\left(2y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-3\right)^2+\left(2y-1\right)^2\ge0\forall x;y\)

Dấu  "=" xảy ra <=> \(\hept{\begin{cases}x-3=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)

Vậy x = 3 ; y = 1/2 

ĐK : 51x \(\ge0\Rightarrow x\ge0\)

Với \(x\ge0\)thì \(x+\frac{1}{1.3}>0;x+\frac{1}{3.5}>0;...;x+\frac{1}{99.101}>0\)

Khi đó : \(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+\left|x+\frac{1}{5.7}\right|+...+\left|x+\frac{1}{99.101}\right|=51x\)

<=> \(x+\frac{1}{1.3}+x+\frac{1}{3.5}+x+\frac{1}{5.7}+....+x+\frac{1}{99.101}=51x\)(50 hạng tử x ở VT)

<=> \(50x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=51x\)

<=> \(x=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\right)\)

<=> \(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

<=> \(x=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{50}{101}\)

Vậy x = 50/101