The second equation give  \(y=\dfrac{3x}{4}\), the thirst give  \(z=\dfrac{3x}{2}\). Subsitute these values for the first equation we have    \(x+\dfrac{3x}{4}-\dfrac{3x}{2}=4\Leftrightarrow x=16\). So  \(x=16,y=12,z=24\)

Add all equations we have  \(3\left(r+s+t+u\right)=30\Rightarrow r+s+t+u=10\).  Compare this equation with the second equation we deduce that  \(t=r+s\). Substitute \(t=r+s\) for the thirst equation we get   \(r+3s=7\).

By the first equation we have  \(\left\{{}\begin{matrix}2r+s=4\\r+3s=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}r=1\\s=2\end{matrix}\right.\). Substitute \(r=1\) for  the fourth equation we deduce that \(u=4\), and the second equation give us   \(t=3\). So  \(r=1,s=2,t=3,u=4\).

Multiply the first equation with 2, the second with 1, the thirst with -1, the fourth with -1, then add all, we have   \(3a+2b=30+30-5-0=55\). Therefore    \(6a+4b=110\).

Put    \(x=123456\),so \(123456^2-123457.123455=x^2-\left(x+1\right)\left(x-1\right)=x^2-\left(x^2-1\right)=1\).

a) \(3^x=243=3^5\Leftrightarrow x=5\).

b) \(5^m=625\Leftrightarrow5^m=5^4\Leftrightarrow m=4\).

c) By  \(16^x.5^{2x}=16^x.25^x=\left(16.25\right)^x=400^x\), so the done equation equivalant to   

                                                          \(400^x=400^1\Leftrightarrow x=1\).

d) \(2^{2n+2}+2^{2n}=160\Leftrightarrow2^{2n}\left(2^2+1\right)=160\Leftrightarrow2^{2n}=2^5\Leftrightarrow2n=5\Leftrightarrow n=2,5.\)

\(m^2+n^2=\left(m+n\right)^2-2mn=103\)

By the assumption  \(\left\{{}\begin{matrix}a+b=4\\a^3+b^3=28\end{matrix}\right.\Rightarrow a^2-ab+b^2=\dfrac{28}{4}=7\Rightarrow2a^2-2ab+2b^2=14\).

Other hand, \(a+b=4\Rightarrow a^2+2ab+b^2=16\). So:

           \(3\left(a^2+b^2\right)=\left(2a^2-2ab+2b^2\right)+\left(a^2+2ab+b^2\right)=14+16\Rightarrow a^2+b^2=10\)

Put   \(a=2^{777},b=3^{555},c=4^{444}\). We have :

\(a=2^{777}=\left(2^7\right)^{111}=128^{111},b=3^{555}=\left(3^5\right)^{111}=243^{111},c=4^{444}=\left(4^4\right)^{111}=256^{111}\Rightarrow a< b< c\).

So:  \(2^{777}< 3^{555}< 4^{444}\).

We have    \(a^m=\dfrac{1}{3},a^n=6\Rightarrow2=6.\dfrac{1}{3}=a^n.a^m=a^{m+n}\), so  

                                                     \(2^{2m+3n}=\left(a^{m+n}\right)^{2m+3n}=a^{\left(m+n\right)\left(2m+3n\right)}\) .

\(27^x.243^y=\left(3^3\right)^x\left(3^5\right)^y=3^{\left(3x+5y\right)}=3^2=9\)

Because (a+1) satisfies 2(x+1)=3(x-1), so that 2(a+1+1)=3a , a = 4.

With a = 4, the equation to solve become  2[3(2+x) - 2(4-x)] = 16 , then x = 2.

It is very easy. the done equation is writend as   \(\left(x-3\right)\left(\dfrac{13}{6}-\dfrac{4}{3}-\dfrac{7}{6}\right)=8\)

\(\Leftrightarrow-\dfrac{1}{3}\left(x-3\right)=8\Leftrightarrow x=-21\) .

We have    \(y=3^2-\left(\dfrac{1}{4}\pi3^2\right)\) and  \(r-3=3\sqrt{2}\Rightarrow r=3+3\sqrt{2}\).

We have  \(AB=AM+BM=10+6=16;BC=BN+CN=6+x;CA=CP+AP=x+10\).

The triangle ABC has his edges   \(a=10+6=16;b=6+x;c=x+10\) . Call  \(p\)  is the halp perimeter of \(ABC\) then  \(p-a=10,p-b=6,p-c=x\) and the area of ABC is

                                      \(\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=\sqrt{60x\left(16+x\right)}\) .

By the assumptions we have    \(60x\left(x+16\right)=\left(120\sqrt{3}\right)^2\Leftrightarrow x^2+16x-720=0\Leftrightarrow x\in\left\{20;-36\right\}\).

So, \(x=20;p=36\), the perimeter of  ABC is  72.

Suppose CH is the height of CAB, then \(CH=\dfrac{2.120\sqrt{3}}{16}=15.\)  The triangle CHA has

                                  \(CHA=90^0;CH=15;CA=30\Rightarrow A=60^0\). 

The measure of angle A is    \(60^0\) .

The circle of radius 3 have an area \(9\pi\). We sign r as the radius of  the pictured quadrant of the done cỉcle, then

\(r=3\sqrt{2}+3\). Put x is the area to calculate, we have  

               \(\dfrac{1}{4}\pi r^2=2x+\pi.3^2+\left(3^2-\dfrac{1}{4}.\pi.3^2\right)=2x+9\left(1+\dfrac{3\pi}{4}\right)\)

      ​ ​\(\Leftrightarrow\dfrac{\pi}{4}\left(3\sqrt{2}+3\right)^2=2x+9\left(1+\dfrac{3\pi}{4}\right)\Leftrightarrow2x=\dfrac{\left(27+18\sqrt{2}\right)\pi}{4}-\dfrac{36+27\pi}{4}\)

       \(\Leftrightarrow x=\dfrac{9\sqrt{2}\pi-36}{4}\)

With the condition \(x\ne0\), \(f\left(x^{-1}\right)=\dfrac{2x^{-1}+3}{2-3x^{-1}}=\dfrac{2+3x}{2x-3}\). The done equation become

                                             \(\dfrac{2-3x}{2x+3}=\dfrac{2\left(2+3x\right)}{2x-3}\) .

With the additional condition \(x\ne\pm\dfrac{3}{2}\) , the above equation equivalent to  

                        \(2\left(2+3x\right)\left(2x+3\right)=\left(2-3x\right)\left(2x-3\right)\)\(\Leftrightarrow18x^2-13x+18=0\)

This equation has no solution.

The volume is   \(2.3.11=66\left(in^3\right)\).

Put  \(t=x^2+18x-24\Leftrightarrow x^2+18x-24-t=0\Leftrightarrow x=-9\pm\sqrt{105+t},\left(t\ge-105\right)\), the equation become    

                                                  \(\dfrac{1}{t+4}+\dfrac{1}{t}=\dfrac{-12}{t-5}\)  (1)

With the condition  \(t\notin\left\{0;-4;5\right\}\), \(\left(1\right)\Leftrightarrow7t^2+21t-10=0\Leftrightarrow t=\dfrac{-21\pm\sqrt{721}}{14}\)(both of these solutions are not smaller  -105). So, the done equation have 4 solutions

                 \(x=-9\pm\pm\sqrt{\dfrac{1449+\sqrt{721}}{14}};x=-9\pm\sqrt{\dfrac{1449-\sqrt{721}}{14}}\)  .

​ ​

Sign \(A=\left(9-3z+z^2\right)\left(9+3z+z^2\right)\), we have \(A=\left(z^2+9\right)^2-\left(3z\right)^2=z^4+9z^2+9^2\).

Therefore                \(\left(z^2-9\right)A=\left(z^2-9\right)\left(z^4+9z^2+9^2\right)=\)

                                                    \(=\left(z^2\right)^3-\left(9\right)^3=\left(z^3\right)^2-\left(3^2\right)^3=\left(27\right)^2-3^6=\left(3^3\right)^2-3^6=0\).

So that  \(A=0\)

So that  

Consider the function  \(f\left(x\right)=x^4+8x^3+19x^2-33x-90\).

We have                                  \(f'\left(x\right)=4x^3+24x^2+38x-33>0,\forall x>1\) .

Therefore \(f\left(x\right)\) is a function increases in \(\left(1;+\infty\right)\). Note that   \(f\left(2\right)=0\), so that in \(\left(1;+\infty\right)\), the done equation has a only soluton \(x=2.\) However  \(f\left(0\right)=-90,f\left(1\right)=-95\). So the done equation has only natural solution \(n=2\).