جزءصحیح (براکت)

$\begin{array}{l}\Rightarrow \left(\left[x\right]+1\right)+\left(\left[-x\right]+4\right)=\left[x\right]+2\end{array}$

$\begin{array}{l}\\ \Rightarrow \left[-x\right]=-3\end{array}$

$\begin{array}{l}\Rightarrow -3\le -x<-2\\ \\ \Rightarrow 2<x\le 3\end{array}$

$\begin{array}{l}\left[2x+\frac{1}{5}\right]=1\\ \\ \Rightarrow 1\le 2x+\frac{1}{5}<2\\ \end{array}$

$\begin{array}{l}\Rightarrow 1-\frac{1}{5}\le 2x<2-\frac{1}{5}\\ \\ \Rightarrow \frac{4}{5}\le 2x<\frac{9}{5}\\ \\ \Rightarrow \frac{2}{5}\le x<\frac{9}{10}\end{array}$

$\begin{array}{l}\forall n\in \mathbb{Z};\left[x+n\right]=\left[x\right]+n\\ \end{array}$

$\begin{array}{l}\\ \left[x+3\right]+\left[-x+2\right]+\left[x-2\right]=3\end{array}$

$\begin{array}{l}\\ \Rightarrow \left(\left[x\right]+3\right)+\left(\left[-x\right]+2\right)+\left(\left[x\right]-2\right)=3\end{array}$

$\begin{array}{l}\Rightarrow \left(\left[x\right]+\left[-x\right]\right)+\left[x\right]=0\\ \end{array}$

$\begin{array}{l}ifx\in \mathbb{Z}\stackrel{\left[x\right]+\left[-x\right]=0}{\to }0+\left[x\right]=0\Rightarrow \left[x\right]=0\Rightarrow x=0\end{array}$

$\begin{array}{l}\\ ifx\notin \mathbb{Z}\stackrel{\left[x\right]+\left[-x\right]=-1}{\to }\left(-1\right)+\left[x\right]=0\Rightarrow \left[x\right]=1\Rightarrow 1<x<2\end{array}$

$D=\left\{0\right\}\cup \left(1,2\right)$

$x^2-4\left[x\right]+4=0\Rightarrow x^2+4=4\left[x\right]\stackrel{n=\left[x\right]}{\to }{x}^2+4=4n$

$\begin{array}{l}n\le x<n+1\\ \\ \Rightarrow n^2\le x^2<{\left(n+1\right)}^2\\ \\ \Rightarrow n^2+4\le x^2+4<{\left(n+1\right)}^2+4\\ \\ \Rightarrow n^2+4\le 4n<{\left(n+1\right)}^2+4\\ \end{array}$

$\begin{array}{l}{n}^2+4\le 4n\Rightarrow n^2-4n+4\le 0\Rightarrow {\left(n-2\right)}^2\le 0\Rightarrow {\left(n-2\right)}^2=0\Rightarrow n-2=0\Rightarrow n=2\end{array}$

$\begin{array}{l}\\ 4n<{\left(n+1\right)}^2+4\Rightarrow 4n<n^2+2n+5\Rightarrow n^2-2n+5>0\Rightarrow n\in R\end{array}$

$D=R\cap \left\{2\right\}=\left\{2\right\}$

$\begin{array}{l}{x}^3-\left[x\right]=3;\left[x\right]=n,n\in \mathbb{Z}\\ \\ \Rightarrow x^3-n=3\\ \\ \Rightarrow x^3=n+3\\ \\ \Rightarrow x=\sqrt[3]{n+3}\\ \end{array}$

$\begin{array}{l}\Rightarrow \left[x\right]=\left[\sqrt[3]{n+3}\right];\left[x\right]=n\\ \\ \Rightarrow n=\sqrt[3]{n+3}\\ \\ \Rightarrow n\le \sqrt[3]{n+3}<n+1\\ \\ \Rightarrow n^3\le \left(n+3\right)<{\left(n+1\right)}^3\end{array}$

مشاهده می‌كنيم كه اين نامساوی فقط به ازای $n=1$ برقرار است:

$\begin{array}{l}n=1\Rightarrow x^3=1+3\Rightarrow x^3=4\Rightarrow x=\sqrt[3]{4}\end{array}$

$\begin{array}{l}\\ D=\left\{\sqrt[3]{4}\right\}\end{array}$

معادله جواب ندارد زيرا طبق تعريف سمت چپ معادله باید یک عدد صحیح باشد اما سمت راست معادله، صحیح نیست.

$D=\varnothing$

$\begin{array}{l}\Rightarrow \left[x\right]+3\left[x\right]=2\left(\left[x\right]-4\right)\\ \\ \Rightarrow 4\left[x\right]=2\left[x\right]-8\\ \end{array}$

$\begin{array}{l}\Rightarrow 2\left[x\right]=-8\\ \\ \Rightarrow \left[x\right]=-4\\ \end{array}$

$\begin{array}{l}\Rightarrow -4\le x<-3\\ \\ \\ D=\left[-4,-3\right)\end{array}$

یادآوری)

$\left[2x\right]=\left[x\right]+\left[x+\frac{1}{2}\right]\Rightarrow \left[2x\right]-\left[x\right]=\left[x+\frac{1}{2}\right]$

$\begin{array}{l}\left[2x\right]-\left[x\right]=2\\ \\ \Rightarrow \left[x+\frac{1}{2}\right]=2\\ \\ \Rightarrow 2\le x+\frac{1}{2}<3\\ \end{array}$

$\begin{array}{l}\Rightarrow \frac{3}{2}\le x<\frac{5}{2}\\ \\ \\ D=\left[\frac{3}{2},\frac{5}{2}\right)\end{array}$

$\begin{array}{l}\Rightarrow \left[x\right]=\frac{5}{6}\left(\left[x\right]+p\right)\\ \\ \Rightarrow \left[x\right]=\frac{5}{6}\left[x\right]+\frac{5}{6}p\\ \\ \Rightarrow \frac{5}{6}p=\left[x\right]-\frac{5}{6}\left[x\right]\\ \end{array}$

$\begin{array}{l}\Rightarrow \frac{5}{6}P=\frac{\left[x\right]}{6}\\ \\ \Rightarrow p=\frac{\left[x\right]}{5};\left(0\le p<1\right)\\ \\ \Rightarrow 0\le \left[x\right]<5\\ \\ \Rightarrow \left[x\right]=0,1,2,3,4\end{array}$

$\begin{array}{l}\left[x\right]=0\Rightarrow 0=\frac{5}{6}x\Rightarrow x=0\\ \\ \left[x\right]=1\Rightarrow 1=\frac{5}{6}x\Rightarrow x=\frac{6}{5}\\ \end{array}$

$\begin{array}{l}\left[x\right]=2\Rightarrow 2=\frac{5}{6}x\Rightarrow x=\frac{12}{5}\\ \\ \left[x\right]=3\Rightarrow 3=\frac{5}{6}x\Rightarrow x=\frac{18}{5}\\ \end{array}$

$\begin{array}{l}\left[x\right]=4\Rightarrow 4=\frac{5}{6}x\Rightarrow x=\frac{24}{5}\\ \\ \\ D=\left\{0,\frac{6}{5},\frac{12}{5},\frac{18}{5},\frac{24}{5}\right\}\end{array}$

$\begin{array}{l}if\left[x\left[x\right]\right]=1\Rightarrow 1\le x\left[x\right]<2\end{array}$

$ifx=n+p\Rightarrow \left\{\begin{array}{l}n=\left[x\right]\\ p=\left\{x\right\}\end{array}\right.〉1\le \left(n+p\right)n<2\Rightarrow 1\le n^2+pn<2;\left({\rm I}\right)$

نامعادله $\left({\rm I}\right)$ برای $n=1$ و $0\le p<1$ برقرار است:

$if0\le p<1,n=1\Rightarrow n\le n+p<n+1\stackrel{n=1}{\to }1\le x<2$

نامعادله $\left({\rm I}\right)$ برای $n\ge 2$ و $n=0$ $n\le -2$ برقرار نیست.

نامعادله $\left({\rm I}\right)$ برای $n=-1$ تنها به‌ازای $p=0$ برقرار است:

$\begin{array}{l}ifn=-1,p=0\Rightarrow x=\left(-1\right)+0\Rightarrow x=-1\end{array}$

$\begin{array}{l}\\ D=\left[1,2\right)\cup \left\{-1\right\}\end{array}$

$\begin{array}{l}\Rightarrow 3\le {10}^x<4\\ \\ \Rightarrow \left\{\begin{array}{l}3\le {10}^x\Rightarrow {\log }_{10}^{}3\le x\\ 4>{10}^x\Rightarrow {\log }_{10}^{}4>x\end{array}\right.\end{array}$

$D=\left(-\infty ,{\log }_{10}^{}4\right)\cap \left[{\log }_{10}^{}3,+\infty \right)=\left[{\log }_{10}^{}3,{\log }_{10}^{}4\right)$

یادآوری)

$\begin{array}{l}{\log }_a^{}N=x\iff N=a^x\end{array}$

$\begin{array}{l}\log \left[x\right]=1\\ \\ \Rightarrow \left[x\right]={10}^1\\ \\ \Rightarrow 10\le x<11\\ \\ D=\left[10,11\right)\end{array}$

$\begin{array}{l}{4}^{\left[x\right]}-6\times 2^{\left[x\right]}+8=0\end{array}$

$\begin{array}{l}\\ \Rightarrow 2^{2\left[x\right]}-6\times 2^{\left[x\right]}+8=0;\left(if{2}^{\left[x\right]}=y\right)\end{array}$

$\begin{array}{l}\Rightarrow y^2-6y+8=0\\ \\ \Rightarrow \left(y-2\right)\left(y-4\right)=0\end{array}$

$\begin{array}{l}ify-2=0\\ \\ \Rightarrow y=2;\left(y=2^{\left[x\right]}\right)\\ \\ \Rightarrow 2^{\left[x\right]}=2^1\\ \end{array}$

$\begin{array}{l}\Rightarrow \left[x\right]=1\\ \\ \Rightarrow 1\le x<2\\ \\ \Rightarrow x\in \left[1,2\right)\end{array}$

$\begin{array}{l}ify-4=0\\ \\ \Rightarrow y=4;\left(y=2^{\left[x\right]}\right)\\ \\ \Rightarrow 2^{\left[x\right]}=4\\ \end{array}$

$\begin{array}{l}\Rightarrow 2^{\left[x\right]}=2^2\\ \\ \Rightarrow \left[x\right]=2\\ \\ \Rightarrow 2\le x<3\\ \\ \Rightarrow x\in \left[2,3\right)\end{array}$

$D=\left[1,2\right)\cup \left[2,3\right)=\left[1,3\right)$

$\begin{array}{l}-1\le sinx\le 1:\\ \\ -1\le \sin x<0\Rightarrow \left[\sin x\right]=-1\\ \\ 0\le \sin x<1\Rightarrow \left[\sin x\right]=0\\ \\ \sin x=1\Rightarrow \left[\sin x\right]=1\end{array}$

$\begin{array}{l}-1\le cosx\le 1:\\ \\ -1\le \cos x<0\Rightarrow \left[\cos x\right]=-1\\ \\ 0\le \cos x<1\Rightarrow \left[\cos x\right]=0\\ \\ \cos x=1\Rightarrow \left[\cos x\right]=1\end{array}$

حالت اول)

$\left\{\begin{array}{l}\left[\sin x\right]=-1\Rightarrow -1\le \sin x<0\\ \left[\cos x\right]=-1\Rightarrow -1\le \cos x<0\end{array}\right.〉x\in \left(2k\pi +\pi ,2k\pi +\frac{3\pi }{2}\right)$

حالت دوم)

$\left\{\begin{array}{l}\left[\sin x\right]=0\Rightarrow 0\le \sin x<1\\ \left[\cos x\right]=0\Rightarrow 0\le \cos x<1\end{array}\right.〉x\in \left(2k\pi ,2k\pi +\frac{\pi }{2}\right)$

حالت سوم)

$\left\{\begin{array}{l}\left[\sin x\right]=1\Rightarrow \sin x=1\\ \left[\cos x\right]=1\Rightarrow \cos x=1\end{array}\right.$

به‌ازای هيچ $x$ ی ،  $\sin x$ و $\cos x$ با هم برابر واحد نمی‌شوند، پس جواب ندارد.

$D=\left(2k\pi ,2k\pi +\frac{\pi }{2}\right)\cup \left(2k\pi +\pi ,2k\pi +\frac{3\pi }{2}\right);k\in Z$

$\begin{array}{l}\left[\sin x+\cos x\right]=-2\\ \\ \Rightarrow -2\le \sin x+\cos x<-1\\ \end{array}$

$\begin{array}{l}\Rightarrow 2k\pi +\pi <x<2k\pi +\frac{3\pi }{2};k\in \mathbb{Z}\end{array}$

$\begin{array}{l}\\ D=\left(2k\pi +\pi ,2k\pi +\frac{3\pi }{2}\right)\end{array}$

$ifx=n+p,n\in \mathbb{Z},0\le p<1$

$\begin{array}{l}\left[x+\frac{3}{8}\right]+\left[x\right]=\frac{7x-2}{3}\end{array}$

$\begin{array}{l}\\ \Rightarrow \left[\left(n+p\right)+\frac{3}{8}\right]+\left[n+p\right]=\frac{7\left(n+p\right)-2}{3}\end{array}$

$\begin{array}{l}\\ \Rightarrow n+\left[p+\frac{3}{8}\right]+n+\left[p\right]=\frac{7n+7p-2}{3};\left[p\right]=0\end{array}$

$\begin{array}{l}\Rightarrow 2n+\left[p+\frac{3}{8}\right]=\frac{7n+7p-2}{3}\\ \end{array}$

$\begin{array}{l}\Rightarrow 3\left(2n+\left[p+\frac{3}{8}\right]\right)=7n+7p-2\\ \\ \Rightarrow 6n+3\left[p+\frac{3}{8}\right]=7n+7p-2\\ \\ \Rightarrow n=3\left[p+\frac{3}{8}\right]-7p+2\end{array}$

$\left.1\right)if0\le p<\frac{5}{8}\Rightarrow \left[p+\frac{3}{8}\right]=0$

$\begin{array}{l}n=3\left[p+\frac{3}{8}\right]-7p+2;\left[p+\frac{3}{8}\right]=0\end{array}$

$\begin{array}{l}\\ \Rightarrow n=-7p+2\\ \\ \Rightarrow p=\frac{n-2}{-7};0\le p<\frac{5}{8}\\ \end{array}$

$\begin{array}{l}\Rightarrow 0\le \frac{n-2}{-7}<\frac{5}{8}\\ \\ \Rightarrow \frac{-7\times 5}{8}<n-2\le 0\\ \\ \Rightarrow \frac{-35}{8}+2<n\le 2\\ \end{array}$

$\begin{array}{l}\Rightarrow \frac{-19}{8}<n\le 2\end{array}$

$\begin{array}{l}\\ \Rightarrow n=-2,-1,0,1,2;p=\frac{n-2}{-7}\end{array}$

$\begin{array}{l}\Rightarrow p=\frac{4}{7},\frac{3}{7},\frac{2}{7},\frac{1}{7},0\\ \\ \\ x=n+p\\ \end{array}$

$x=\left(-2+\frac{4}{7}\right),\left(-1+\frac{3}{7}\right),\left(0+\frac{2}{7}\right),\left(1+\frac{1}{7}\right),\left(2+0\right)$

برای $x$ پنج جواب به‌صورت فوق به‌دست می‌آید.

$\left.2\right)ifp\ge \frac{5}{8}$

به‌همین ترتیب مانند قسمت اول، برای حالت $p\ge \frac{5}{8}$ هم دو جواب به‌دست می‌آید.

$D_f:\left\{\begin{array}{l}x+\left[x\right]\ge 0\Rightarrow \left[x\right]\ge -x\Rightarrow x\in \left[0,+\infty \right)\\ x-\left[x\right]\ge 0\Rightarrow \left[x\right]\le x\Rightarrow x\in R\end{array}\right.〉x\in \left[0,+\infty \right)$

به‌ازای هیچ مقداری از دامنه، معادله فوق برقرار نیست، پس معادله جواب ندارد.

$\begin{array}{l}nx-\left(n+1\right)\left[x\right]=0;x=\left[x\right]+p\\ \\ \Rightarrow n\left(\left[x\right]+p\right)-\left(n+1\right)\left[x\right]=0\\ \end{array}$

$\begin{array}{l}\Rightarrow n\left[x\right]+np-\left(n+1\right)\left[x\right]=0\\ \\ \Rightarrow n\left[x\right]+np-n\left[x\right]-\left[x\right]=0\\ \\ \Rightarrow np=\left[x\right]\end{array}$

$\begin{array}{l}0\le p<1\Rightarrow 0\le np<n\stackrel{np=\left[x\right]}{\to }0\le \left[x\right]<n\end{array}$

$\begin{array}{l}\\ if\left[x\right]=k,k\in \mathbb{Z}\Rightarrow k=0,1,2,\cdot \cdot \cdot ,n-1\end{array}$

$\begin{array}{l}\\ ifnp=\left[x\right]\Rightarrow p=\frac{\left[x\right]}{n}\Rightarrow p=\frac{k}{n}\end{array}$

$\begin{array}{l}\\ ifx=\left[x\right]+p\Rightarrow x=k+\frac{k}{n}\Rightarrow x=k\left(\frac{n+1}{n}\right)\end{array}$

معادله زیر $n$ ریشه دارد:

$\begin{array}{l}k=0,1,2,\cdot \cdot \cdot ,n-1\\ \\ x=k\left(\frac{n+1}{n}\right)\end{array}$

$\frac{\left[x\right]}{4}=k;x\in Z\Rightarrow \left[x\right]=4k\Rightarrow 4k\le x<4k+1$

یادآوری)

$\begin{array}{l}\left[2x\right]=\left[x\right]+\left[x+\frac{1}{2}\right]\end{array}$

$\begin{array}{l}\left[2x\right]=\left[x\right]\\ \\ \Rightarrow \left[x\right]+\left[x+\frac{1}{2}\right]=\left[x\right]\\ \\ \Rightarrow \left[x+\frac{1}{2}\right]=0\\ \end{array}$

$\begin{array}{l}\Rightarrow 0\le x+\frac{1}{2}<1\\ \\ \Rightarrow -\frac{1}{2}\le x<\frac{1}{2}\end{array}$