Problem 1
Find all the roots of equation \(2x-2.8=\lfloor x\rfloor\).
Solution 1
\[\begin{align}
2x-2.8=\lfloor x\rfloor&\Longrightarrow x-1<2x-2.8\le x\\
&\Longleftrightarrow1.8<x\le2.8\\
&\Longrightarrow\lfloor x\rfloor=1\lor\lfloor x\rfloor=2
\end{align}
\]
If \(\lfloor x\rfloor=1\),
\[\begin{align}
2x-2.8=\lfloor x\rfloor&\Longleftrightarrow2x-2.8=1\\
&\Longleftrightarrow x=1.9
\end{align}
\]
If \(\lfloor x\rfloor=2\),
\[\begin{align}
2x-2.8=\lfloor x\rfloor&\Longleftrightarrow2x-2.8=2\\
&\Longleftrightarrow x=2.4
\end{align}
\]
Therefore,
\[2x-2.8=\lfloor x\rfloor\Longleftrightarrow x=1.9\lor x=2.4
\]
The roots of equation \(2x-2.8=\lfloor x\rfloor\) are \(x=1.9\) and \(x=2.4\).
Problem 2
Solve equation \(x^2-\lfloor x\rfloor-2=0\).
Solution 2
\[\begin{align}
x^2-\lfloor x\rfloor-2=0&\Longleftrightarrow x^2-2=\lfloor x\rfloor\\
&\Longrightarrow x-1<x^2-2\le x\\
&\Longleftrightarrow x-1<x^2-2\land x^2-2\le x\\
&\Longleftrightarrow(x<\frac{1-\sqrt5}{2}\lor x>\frac{1+\sqrt5}{2})\land(-1\le x\le2)\\
&\Longleftrightarrow-1\le x<\frac{1-\sqrt5}{2}\lor\frac{1+\sqrt5}{2}<x\le2\\
&\Longrightarrow\lfloor x\rfloor=-1\lor\lfloor x\rfloor=1\lor\lfloor x\rfloor=2
\end{align}
\]
If \(\lfloor x\rfloor=-1\),
\[\begin{align}
x^2-2=\lfloor x\rfloor&\Longleftrightarrow x^2-2=-1\\
&\Longleftrightarrow x=-1
\end{align}
\]
If \(\lfloor x\rfloor=1\),
\[\begin{align}
x^2-2=\lfloor x\rfloor&\Longleftrightarrow x^2-2=1\\
&\Longleftrightarrow x=\sqrt3
\end{align}
\]
If \(\lfloor x\rfloor=2\),
\[\begin{align}
x^2-2=\lfloor x\rfloor&\Longleftrightarrow x^2-2=2\\
&\Longleftrightarrow x=2
\end{align}
\]
Therefore,
\[x^2-\lfloor x\rfloor-2=0\Longleftrightarrow x=-1\lor x=\sqrt3\lor x=2
\]
The roots of equation \(x^2-\lfloor x\rfloor-2=0\) are \(x=-1\), \(x=\sqrt3\) and \(x=2\).
Problem 3
Solve equation \(\lfloor\frac{5x-13}{3}\rfloor=\frac{3x+1}{5}\).
Solution 3
\[\begin{align}
\left\lfloor\frac{5x-13}{3}\right\rfloor=\frac{3x+1}{5}&\Longleftrightarrow\exist k\in\Z,\left\lfloor\frac{5x-13}{3}\right\rfloor=k\land\frac{3x+1}{5}=k\\
&\Longleftrightarrow\exist k\in\Z,\left\lfloor\frac{5x-13}{3}\right\rfloor=k\land x=\frac{5k-1}{3}\\
&\Longleftrightarrow\exist k\in\Z,\left\lfloor\frac{25k-44}{9}\right\rfloor=k\land x=\frac{5k-1}{3}\\
\end{align}
\]
\[\begin{align}
\left\lfloor\frac{25k-44}{9}\right\rfloor=k&\Longleftrightarrow k\le\frac{25k-44}{9}<k+1\\
&\Longleftrightarrow\frac{11}{4}\le k<\frac{53}{16}\\
&\Longleftrightarrow k=3
\end{align}
\]
Therefore,
\[\begin{align}
\left\lfloor\frac{5x-13}{3}\right\rfloor=\frac{3x+1}{5}&\Longleftrightarrow\exist k\in\Z,k=3\land x=\frac{5k-1}{3}\\
&\Longleftrightarrow x=\frac{14}{3}
\end{align}
\]
The root of equation \(\lfloor\frac{5x-13}{3}\rfloor=\frac{3x+1}{5}\) is \(x=\frac{14}{3}\).
Problem 4
\[3x-3=\lfloor x\rfloor
\]
Solution 4
\[\begin{align}
3x-3=\lfloor x\rfloor&\Longrightarrow x-1<3x-3\le x\\
&\Longleftrightarrow1<x\le\frac{3}{2}\\
&\Longrightarrow\lfloor x\rfloor=1\\
\end{align}
\]
If \(\lfloor x\rfloor=1\),
\[\begin{align}
3x-3=\lfloor x\rfloor&\Longleftrightarrow3x-3=1\\
&\Longleftrightarrow x=\frac{4}{3}
\end{align}
\]
Therefore,
\[3x-3=\lfloor x\rfloor\Longleftrightarrow x=\frac{4}{3}
\]
Problem 5
\[5-x^2=\lfloor x\rfloor
\]
Solution 5
\[\begin{align}
5-x^2=\lfloor x\rfloor&\Longrightarrow x-1<5-x^2\le x\\
&\Longleftrightarrow x-1<5-x^2\land5-x^2\le x\\
&\Longleftrightarrow-3<x<2\land(x\le\frac{-1-\sqrt{21}}{2}\lor x\ge\frac{-1+\sqrt{21}}{2})\\
&\Longleftrightarrow-3<x\le\frac{-1-\sqrt{21}}{2}\lor\frac{-1+\sqrt{21}}{2}\le x<2\\
&\Longrightarrow\lfloor x\rfloor=-3\lor\lfloor x\rfloor=1
\end{align}
\]
If \(\lfloor x\rfloor=-3\),
\[5-x^2=-3\Longleftrightarrow x=-2\sqrt2
\]
If \(\lfloor x\rfloor=1\),
\[5-x^2=1
\]
Therefore,
\[5-x^2=\lfloor x\rfloor\Longleftrightarrow x=-2\sqrt2
\]
Problem 6
\[|x-2|=\lfloor x\rfloor
\]
Solution 6
\[\begin{align}
\lfloor x\rfloor\in\Z&\Longleftrightarrow|x-2|\in\Z\\
&\Longleftrightarrow x\in\Z\\
&\Longleftrightarrow\lfloor x\rfloor=x\\
\end{align}
\]
Therefore,
\[\begin{align}
|x-2|=\lfloor x\rfloor&\Longleftrightarrow|x-2|=x\\
&\Longleftrightarrow x=1
\end{align}
\]
Problem 7
\[\frac{3x+2}{4}=\left\lfloor\frac{6x-1}{5}\right\rfloor
\]
Solution 7
\[\begin{align}
\frac{3x+2}{4}=\left\lfloor\frac{6x-1}{5}\right\rfloor&\Longleftrightarrow\exist k\in\Z,\frac{3x+2}{4}=k\land\left\lfloor\frac{6x-1}{5}\right\rfloor=k\\
&\Longleftrightarrow\exist k\in\Z,x=\frac{4k-2}{3}\land\left\lfloor\frac{6x-1}{5}\right\rfloor=k\\
&\Longleftrightarrow\exist k\in\Z,x=\frac{4k-2}{3}\land\left\lfloor\frac{8k}{5}\right\rfloor=k+1
\end{align}
\]
\[\begin{align}
\left\lfloor\frac{8k}{5}\right\rfloor=k+1&\Longleftrightarrow k+1\le\frac{8k}{5}<k+2\\
&\Longleftrightarrow k+1\le\frac{8k}{5}<k+2\\
&\Longleftrightarrow\frac{5}{3}\le k<\frac{10}{3}\\
&\Longleftrightarrow k=2\lor k=3
\end{align}
\]
Therefore,
\[\begin{align}
\frac{3x+2}{4}=\left\lfloor\frac{6x-1}{5}\right\rfloor&\Longleftrightarrow\exist k\in\Z,x=\frac{4k-2}{3}\land(k=2\lor k=3)\\
&\Longleftrightarrow x=2\lor x=\frac{10}{3}
\end{align}
\]
Problem 8
\[\begin{cases}
y=2\lfloor x\rfloor+\sqrt2\\
x=3\lfloor y\rfloor-12.5
\end{cases}
\]
Solution 8
\[\begin{align}
y=2\lfloor x\rfloor+\sqrt2\land x=3\lfloor y\rfloor-12.5&\Longleftrightarrow\exist k\in\Z,\lfloor x\rfloor=k\land y=2k+\sqrt2\land x=3\lfloor y\rfloor-12.5\\
&\Longleftrightarrow\exist k\in\Z,\lfloor x\rfloor=k\land y=2k+\sqrt2\land x=3\lfloor2k+\sqrt2\rfloor-12.5\\
&\Longleftrightarrow\exist k\in\Z,\lfloor x\rfloor=k\land y=2k+\sqrt2\land x=6k-9.5\\
&\Longleftrightarrow\exist k\in\Z,\lfloor6k-9.5\rfloor=k\land y=2k+\sqrt2\land x=6k-9.5\\
\end{align}
\]
\[\begin{align}
\lfloor6k-9.5\rfloor=k&\Longleftrightarrow k\le6k-9.5<k+1\\
&\Longleftrightarrow1.9\le k<2.1\\
&\Longleftrightarrow k=2\\
\end{align}
\]
Therefore,
\[\begin{align}
y=2\lfloor x\rfloor+\sqrt2\land x=3\lfloor y\rfloor-12.5&\Longleftrightarrow\exist k\in\Z,k=2\land y=2k+\sqrt2\land x=6k-9.5\\
&\Longleftrightarrow x=2.5\land y=4+\sqrt2
\end{align}
\]