Korzystamy ze wzorów:

$a^m\cdot a^n=a^{m+n}$

$\frac{b^m}{b^n}=b^{m-n}$

 

${\left(-1\frac{1}{2}\right)}^7\cdot \left(-1\frac{1}{2}\right):{\left(-1\frac{1}{2}\right)}^6={\left(-1\frac{1}{2}\right)}^{7+1}:{\left(-1\frac{1}{2}\right)}^6={\left(-1\frac{1}{2}\right)}^8:{\left(-1\frac{1}{2}\right)}^6={\left(-1\frac{1}{2}\right)}^{8-6}=$

$={\left(-1\frac{1}{2}\right)}^2={\left(-\frac{3}{2}\right)}^2=\frac{9}{4}=2\frac{1}{4}$

$2\frac{1}{4}---\to T$

 

$\underline{}$

$\frac{{\left(\frac{1}{4}\right)}^8\cdot 8^4}{{\left(\frac{1}{4}\right)}^7\cdot 8^3}=\frac{{\left(\frac{1}{4}\right)}^8}{{\left(\frac{1}{4}\right)}^7}\cdot \frac{8^4}{8^3}={\left(\frac{1}{4}\right)}^{8-7}\cdot 8^{4-3}={\left(\frac{1}{4}\right)}^1\cdot 8^1=\frac{1}{{\sout{4}}^1}\cdot {\sout{8}}^2=2$

$2---\to L$

 

$\underline{}$

$\frac{{\left(-0,1\right)}^7\cdot {\left(-0,1\right)}^2}{{\left(0,1\right)}^5\cdot {\left(0,1\right)}^4}=\frac{{\left(-0,1\right)}^9}{{\left(0,1\right)}^9}=\frac{-{\sout{0,1}}^{\sout{9}}}{{\sout{0,1}}^{\sout{9}}}=-1$

$-1---\to O$

 

$\underline{}$

$\frac{{\left(-0,2\right)}^3\cdot {\left(-0,2\right)}^4\cdot {\left(-0,2\right)}^5}{{\left(-0,2\right)}^8\cdot {\left(-0,2\right)}^4}=\frac{{\left(-0,2\right)}^{12}}{{\left(-0,2\right)}^{12}}={\left(-0,2\right)}^0=1$

$1---\to I$

 

$\underline{}$

$\frac{5^8\cdot {\left(0,1\right)}^7}{5^6\cdot {\left(0,1\right)}^4}=5^{8-6}\cdot {\left(0,1\right)}^{7-4}=5^2\cdot {\left(0,1\right)}^3=25\cdot 0,001={\sout{25}}^1\cdot \frac{1}{{\sout{1000}}^{40}}=\frac{1}{40}$

$\frac{1}{40}---\to E$

 

$\underline{}$

${\left(-1\frac{1}{2}\right)}^5\cdot {\left(-1\frac{1}{2}\right)}^8:{\left(1\frac{1}{2}\right)}^{11}={\left(-1\frac{1}{2}\right)}^{13}:{\left(1\frac{1}{2}\right)}^{11}=$

$-{\left(1\frac{1}{2}\right)}^{13}:{\left(1\frac{1}{2}\right)}^{11}=-{\left(1\frac{1}{2}\right)}^2=-{\left(\frac{3}{2}\right)}^2=-\frac{9}{4}=-2\frac{1}{4}$

$-2\frac{1}{4}---\to W$

 

Tabelka:

-2 1/4	1	2 1/4	1 /40	2	-1
W	I	T	E	L	O