قدرمطلق

$\begin{array}{l}x\left|y\right|+y\left|x\right|=x\left|\frac{1}{x}\right|+\frac{1}{x}\left|x\right|\\ \\ \Rightarrow x\left|y\right|+y\left|x\right|=\frac{x}{\left|x\right|}+\frac{\left|x\right|}{x}\end{array}$

$\begin{array}{l}\Rightarrow x\left|y\right|+y\left|x\right|=\left\{\begin{array}{l}\frac{x}{x}+\frac{x}{x};x>0\\ \frac{x}{-x}+\frac{-x}{x};x<0\end{array}\right.\end{array}$

$\begin{array}{l}\\ \Rightarrow x\left|y\right|+y\left|x\right|=\left\{\begin{array}{l}2;x>0\\ -2;x<0\end{array}\right.\end{array}$

$\begin{array}{l}\sqrt{a^2+1+2\sqrt{a^2}}\\ \\ =\sqrt{a^2+1+2\left|a\right|}\end{array}$

$\begin{array}{l}=\sqrt{a^2+2\left|a\right|+1};a<0\Rightarrow \left|a\right|=-a\end{array}$

$\begin{array}{l}=\sqrt{a^2-2a+1}\\ \\ =\sqrt{{\left(a-1\right)}^2}\end{array}$

$\begin{array}{l}=\left|a-1\right|;ifa<0\Rightarrow a-1<-1\Rightarrow \left|a-1\right|=-(a-1)\end{array}$

$\begin{array}{l}=-\left(a-1\right)\\ \\ =1-a\end{array}$

حالت اول)

$\begin{array}{l}ifa>0,b<0\\ \\ \left(\frac{a}{\left|a\right|}+\frac{b}{\left|b\right|}\right)\left(\frac{a-b}{|a-b|}+\frac{ab}{\left|ab\right|}\right)\end{array}$

$\begin{array}{l}=\left(\frac{a}{a}+\frac{b}{-b}\right)\left(\frac{a-b}{\left|a-b\right|}+\frac{ab}{\left|ab\right|}\right)\end{array}$

$\begin{array}{l}\\ =\left(0\right)\left(\frac{a-b}{\left|a-b\right|}+\frac{ab}{\left|ab\right|}\right)\\ \\ =0\end{array}$

حالت دوم)

$\begin{array}{l}ifa<0,b>0\\ \\ \left(\frac{a}{\left|a\right|}+\frac{b}{\left|b\right|}\right)\left(\frac{a-b}{|a-b|}+\frac{ab}{\left|ab\right|}\right)\end{array}$

$\begin{array}{l}\\ =\left(\frac{a}{-a}+\frac{b}{b}\right)\left(\frac{a-b}{\left|a-b\right|}+\frac{ab}{\left|ab\right|}\right)\end{array}$

$\begin{array}{l}=\left(0\right)\left(\frac{a-b}{\left|a-b\right|}+\frac{ab}{\left|ab\right|}\right)\\ \\ =0\end{array}$

درهر دو حالت، به نتیجه زیر می‌رسیم:

$ifab<0\Rightarrow \left(\frac{a}{\left|a\right|}+\frac{b}{\left|b\right|}\right)\left(\frac{a-b}{\left|a-b\right|}+\frac{ab}{\left|ab\right|}\right)=0$

$\begin{array}{l}\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+3-4\sqrt{x-1}}\end{array}$

$\begin{array}{l}\\ =\sqrt{x-1+4+4\sqrt{x-1}}-\sqrt{x-1+4-4\sqrt{x-1}}\end{array}$

$\begin{array}{l}=\sqrt{{\left(\sqrt{x-1}+2\right)}^2}-\sqrt{{\left(\sqrt{x-1}-2\right)}^2}\end{array}$

$\begin{array}{l}\\ =\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}-2\right|\\ \\ =\sqrt{x-1}+2-\left|\sqrt{x-1}-2\right|;\left({\rm I}\right)\end{array}$

$\begin{array}{l}=\left\{\begin{array}{l}\sqrt{x-1}+2-\left[-\left(\sqrt{x-1}-2\right)\right];1\le x\le 5\\ \sqrt{x-1}+2-\left[+\left(\sqrt{x-1}-2\right)\right];x>5\end{array}\right.\end{array}$

$\begin{array}{l}\\ =\left\{\begin{array}{l}2\sqrt{x-1};1\le x\le 5\\ 4;x>5\end{array}\right.\end{array}$

$\begin{array}{l}if1\le x\le 5\Rightarrow 0\le x-1\le 4\Rightarrow 0\le \sqrt{x-1}\le 2\Rightarrow -2\le \sqrt{x-1}-2\le 0\Rightarrow \left|\sqrt{x-1}-2\right|=-\left(\sqrt{x-1}-2\right)\end{array}$

$\begin{array}{l}\\ ifx>5\Rightarrow x-1>4\Rightarrow \sqrt{x-1}>2\Rightarrow \sqrt{x-1}-2>0\Rightarrow \left|\sqrt{x-1}-2\right|=+\left(\sqrt{x-1}-2\right)\end{array}$

$\begin{array}{l}\sqrt{1+\sin x}+\sqrt{1-\sin x}\end{array}$

$\begin{array}{l}=\sqrt{\left({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}\right)+2\sin \frac{x}{2}.\cos \frac{x}{2}}+\sqrt{\left({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}\right)-2\sin \frac{x}{2}\cos \frac{x}{2}}\end{array}$

$\begin{array}{l}=\sqrt{{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)}^2}+\sqrt{{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)}^2}\end{array}$

$\begin{array}{l}\\ =\left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|+\left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|\end{array}$

$\begin{array}{l}=\sin \frac{x}{2}+\cos \frac{x}{2}+\left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|;\left({\rm I}\right)\end{array}$

$\begin{array}{l}\\ =\sin \frac{x}{2}+\cos \frac{x}{2}-\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)\end{array}$

$\begin{array}{l}=\sin \frac{x}{2}+\cos \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}\end{array}$

$\begin{array}{l}\\ =2\cos \frac{x}{2}\end{array}$

$\left({\rm I}\right):if0<x<\frac{\pi }{2}\Rightarrow 0<\frac{x}{2}<\frac{\pi }{4}\Rightarrow \cos \frac{x}{2}>\sin \frac{x}{2}\Rightarrow \sin \frac{x}{2}-\cos \frac{x}{2}<0\Rightarrow \left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|=-\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)$

$\begin{array}{l}\frac{1}{\sqrt{x+2\sqrt{x-1}}}+\frac{1}{\sqrt{x-2\sqrt{x-1}}}\end{array}$

$\begin{array}{l}\\ =\frac{1}{\sqrt{\left(x-1\right)+1+2\sqrt{x-1}}}+\frac{1}{\sqrt{\left(x-1\right)+1-2\sqrt{x-1}}}\end{array}$

$\begin{array}{l}=\frac{1}{\sqrt{{\left(\sqrt{x-1}+1\right)}^2}}+\frac{1}{\sqrt{{\left(\sqrt{x-1}-1\right)}^2}}\end{array}$

$\begin{array}{l}\\ =\frac{1}{\left|\sqrt{x-1}+1\right|}+\frac{1}{\left|\sqrt{x-1}-1\right|}\end{array}$

$\begin{array}{l}=\frac{1}{\sqrt{x-1}+1}+\frac{1}{\left|\sqrt{x-1}-1\right|};\left({\rm I}\right)\end{array}$

$\begin{array}{l}\\ =\frac{1}{\sqrt{x-1}+1}+\frac{-1}{\sqrt{x-1}-1}\\ \\ =\frac{\left(\sqrt{x-1}-1\right)-\left(\sqrt{x-1}+1\right)}{\left(\sqrt{x-1}+1\right)\left(\sqrt{x-1}-1\right)}\end{array}$

$\begin{array}{l}=\frac{-2}{\left(x-1\right)-1}\\ \\ =\frac{-2}{x-2}\\ \\ =\frac{2}{2-x}\end{array}$

$\left({\rm I}\right):if1<x<2\Rightarrow 0<x-1<1\Rightarrow 0<\sqrt{x-1}<1\Rightarrow -1<\sqrt{x-1}-1<0\Rightarrow \left|\sqrt{x-1}-1\right|=-\left(\sqrt{x-1}-1\right)$

$\begin{array}{l}\frac{x\sqrt{\frac{x^4-2x^2+1}{4x^2}+1}}{x^2+1}\\ \\ =\frac{x\sqrt{\frac{x^4-2x^2+1+4x^2}{4x^2}}}{x^2+1}\end{array}$

$\begin{array}{l}=\frac{x\sqrt{\frac{{\left(x^2+1\right)}^2}{{\left(2x\right)}^2}}}{x^2+1}\\ \\ =\frac{x.\frac{x^2+1}{\left|2x\right|}}{x^2+1}\end{array}$

$\begin{array}{l}=\frac{x}{\left|2x\right|}\\ \\ =\left\{\begin{array}{l}\frac{1}{2};x>0\\ -\frac{1}{2};x<0\end{array}\right.\end{array}$

$\begin{array}{l}\sqrt{x^2-4\sqrt{x^2}+4}+\sqrt{x^2+4\sqrt{x^2}+4}\end{array}$

$\begin{array}{l}\\ =\sqrt{x^2-4\left|x\right|+4}+\sqrt{x^2+4\left|x\right|+4}\end{array}$

$\begin{array}{l}\\ =\sqrt{{\left|x\right|}^2-4\left|x\right|+4}+\sqrt{{\left|x\right|}^2+4\left|x\right|+4}\end{array}$

$\begin{array}{l}=\sqrt{{\left(\left|x\right|-2\right)}^2}+\sqrt{{\left(\left|x\right|+2\right)}^2}\end{array}$

$\begin{array}{l}\\ =\left|\left|x\right|-2\right|+\left|\left|x\right|+2\right|;\left(-2<x<0\Rightarrow \left|x\right|=-x\right)\end{array}$

$\begin{array}{l}=\left|-x-2\right|+\left|-x+2\right|;\left({\rm I}\right)\\ \\ =-\left(-x-2\right)+\left(-x+2\right)\\ \\ =x+2+\left(-x+2\right)\\ \\ =4\end{array}$

$\begin{array}{l}-2<x<0\Rightarrow 0<-x<2\Rightarrow -2<-x-2<0\Rightarrow \left|-x-2\right|=-\left(-x-2\right)\end{array}$

$\begin{array}{l}\\ -2<x<0\Rightarrow 0<-x<2\Rightarrow 2<-x+2<4\Rightarrow \left|-x+2\right|=+\left(-x+2\right)\end{array}$

$\begin{array}{l}\sqrt{a^2-4a+4}+\sqrt{a^2+6a+9}\\ \\ =\sqrt{{\left(a-2\right)}^2}+\sqrt{{\left(a+3\right)}^2}\end{array}$

$\begin{array}{l}=\left|a-2\right|+\left|a+3\right|;\left({\rm I}\right)\\ \\ =\left(a-2\right)+\left(a+3\right)\\ \\ =2a+1\end{array}$

$\left({\rm I}\right):ifa\ge 2\Rightarrow \left\{\begin{array}{l}a-2\ge 0\Rightarrow \left|a-2\right|=+\left(a-2\right)\\ a+3\ge 5\Rightarrow \left|a+3\right|=+\left(a+3\right)\end{array}\right.$

$\begin{array}{l}\sqrt{\frac{1+\sin x}{1-\sin x}}-\sqrt{\frac{1-\sin x}{1+\sin x}}\end{array}$

$\begin{array}{l}=\sqrt{\frac{\left(1+\sin x\right)\left(1+\sin x\right)}{\left(1-\sin x\right)\left(1+\sin x\right)}}-\sqrt{\frac{\left(1-\sin x\right)\left(1-\sin x\right)}{\left(1+\sin x\right)\left(1-\sin x\right)}}\end{array}$

$\begin{array}{l}=\sqrt{\frac{{\left(1+\sin x\right)}^2}{1-{\sin }^{2}x}}-\sqrt{\frac{{\left(1-\sin x\right)}^2}{1-{\sin }^{2}x}}\\ \\ =\sqrt{\frac{{\left(1+\sin x\right)}^2}{{\cos }^{2}x}}-\sqrt{\frac{{\left(1-\sin x\right)}^2}{{\cos }^{2}x}}\end{array}$

$\begin{array}{l}=\frac{\left|1+\sin x\right|}{\left|\cos x\right|}-\frac{\left|1-\sin x\right|}{\left|\cos x\right|};\left({\rm I}\right)\\ \\ =\frac{\left(1+\sin x\right)}{\left|\cos x\right|}-\frac{\left(1-\sin x\right)}{\left|\cos x\right|}\\ \\ =\frac{2\sin x}{\left|\cos x\right|}\end{array}$

$\left({\rm I}\right):-1\le \sin x\le 1\Rightarrow \left\{\begin{array}{l}\sin x+1\ge 0\Rightarrow \left|1+\sin x\right|=+\left(1+\sin x\right)\\ 1-\sin x\ge 0\Rightarrow \left|1-\sin x\right|=+\left(1-\sin x\right)\end{array}\right.$

$\begin{array}{l}\sqrt{\frac{1+\cos x}{1-\cos x}}-\sqrt{\frac{1-\cos x}{1+\cos x}}\end{array}$

$\begin{array}{l}=\sqrt{\frac{\left(1+\cos x\right)\left(1+\cos x\right)}{\left(1-\cos x\right)\left(1+\cos x\right)}}-\sqrt{\frac{\left(1-\cos x\right)\left(1-\cos x\right)}{\left(1+\cos x\right)\left(1-\cos x\right)}}\end{array}$

$\begin{array}{l}=\sqrt{\frac{{\left(1+\cos x\right)}^2}{\left(1-{\cos }^{2}x\right)}}-\sqrt{\frac{{\left(1-\cos x\right)}^2}{\left(1-{\cos }^{2}x\right)}}\\ \\ =\sqrt{\frac{{\left(1+\cos x\right)}^2}{{\sin }^{2}x}}-\sqrt{\frac{{\left(1-\cos x\right)}^2}{{\sin }^{2}x}}\end{array}$

$\begin{array}{l}=\frac{\left|1+\cos x\right|}{\left|\sin x\right|}-\frac{\left|1-\cos x\right|}{\left|\sin x\right|};\left({\rm I}{\rm I}\right)\\ \\ =\frac{\left(1+\cos x\right)}{\left|\sin x\right|}-\frac{\left(1-\cos x\right)}{\left|\sin x\right|}\\ \\ =\frac{2\cos x}{\left|\sin x\right|}\end{array}$

$\left({\rm I}{\rm I}\right):-1\le \cos x\le 1\Rightarrow \left\{\begin{array}{l}\cos x+1\ge 0\Rightarrow \left|1+\cos x\right|=+\left(1+\cos x\right)\\ 1-\cos x\ge 0\Rightarrow \left|1-\cos x\right|=+\left(1-\cos x\right)\end{array}\right.$

$A=\left(\sqrt{\frac{1+\sin x}{1-\sin x}}-\sqrt{\frac{1-\sin x}{1+\sin x}}\right)\left(\sqrt{\frac{1+\cos x}{1-\cos x}}-\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{2\sin x}{\left|\cos x\right|}.\frac{2\cos x}{\left|\sin x\right|}=\frac{4\sin x\cdot \cos x}{\left|\sin x\cdot \cos x\right|}$

وقتی كه انتهای كمان $x$ در ربع اول يا سوم دايره مثلثاتی باشد، سینوس و کسینوس هم علامت هستند در نتيجه حاصل ضرب آنها مثبت است:

$if\left\{\begin{array}{l}2k\pi <x<2k\pi +\frac{\pi }{2}\\ 2k\pi +\pi <x<2k\pi +\frac{3\pi }{2}\end{array}\right.〉A=\frac{4\sin x\cdot \cos x}{\sin x\cdot \cos x}\Rightarrow A=4$

وقتی كه انتهای كمان $x$ در ربع دوم يا چهارم دايره مثلثاتی باشد، سینوس و کسینوس علامت های مختلف دارند يعنی حاصل ضربشان منفی است.

$if\left\{\begin{array}{l}2k\pi +\frac{\pi }{2}<x<2k\pi +\pi \\ 2k\pi +\frac{3\pi }{2}<x<2k\pi +2\pi \end{array}\right.〉A=\frac{4\sin x\cos x}{-\sin x\cos x}=-4$

$\left|x-3\right|=7\Rightarrow \left\{\begin{array}{l}x-3=7\Rightarrow x=10\\ x-3=-7\Rightarrow x=-4\end{array}\right.$

$\begin{array}{l}2\left|x-6\right|=4\\ \\ \Rightarrow \left|x-6\right|=2\\ \\ \Rightarrow \left\{\begin{array}{l}x-6=2\Rightarrow x=8\\ x-6=-2\Rightarrow x=4\end{array}\right.\end{array}$

$\begin{array}{l}\left|x-\left(-3\right)\right|>2\\ \\ \Rightarrow \left|x+3\right|>2\\ \\ \Rightarrow \left\{\begin{array}{l}x+3>2\Rightarrow x>-1\\ x+3<-2\Rightarrow x<-5\end{array}\right.\end{array}$