Kwadrat liczby to inaczej druga potęga liczby.

 

a)

$5^2=5\cdot 5=25$  

${\left(-2\right)}^2=\left(-2\right)\cdot \left(-2\right)=4$  

${\left(-12\right)}^2=\left(-12\right)\cdot \left(-12\right)=144$  

${\left(-3\right)}^2=\left(-3\right)\cdot \left(-3\right)=9$  

${15}^2=15\cdot 15=225$  

 

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b)

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

${\left(\frac{4}{11}\right)}^2=\frac{4}{11}\cdot \frac{4}{11}=\frac{16}{121}$  

${\left(-\frac{6}{7}\right)}^2=\left(-\frac{6}{7}\right)\cdot \left(-\frac{6}{7}\right)=\frac{36}{49}$  

${\left(\frac{11}{13}\right)}^2=\frac{11}{13}\cdot \frac{11}{13}=\frac{121}{169}$  

${\left(-\frac{10}{17}\right)}^2=\left(-\frac{10}{17}\right)\cdot \left(-\frac{10}{17}\right)=\frac{100}{289}$  

 

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c) 

${\left(4\frac{1}{2}\right)}^2={\left(\frac{9}{2}\right)}^2=\frac{9}{2}\cdot \frac{9}{2}=\frac{81}{4}=20\frac{1}{4}$  

${\left(-3\frac{2}{5}\right)}^2={\left(-\frac{17}{5}\right)}^2=\left(-\frac{17}{5}\right)\cdot \left(-\frac{17}{5}\right)=\frac{289}{25}=11\frac{14}{25}$  

${\left(2\frac{1}{9}\right)}^2={\left(\frac{19}{9}\right)}^2=\frac{19}{9}\cdot \frac{19}{9}=\frac{361}{81}=4\frac{37}{81}$  

${\left(-5\frac{1}{7}\right)}^2={\left(-\frac{36}{7}\right)}^2=\left(-\frac{36}{7}\right)\cdot \left(-\frac{36}{7}\right)=\frac{1296}{49}=26\frac{22}{49}$  

${\left(4\frac{2}{3}\right)}^2={\left(\frac{14}{3}\right)}^2=\frac{14}{3}\cdot \frac{14}{3}=\frac{196}{9}=21\frac{7}{9}$  

 

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d)

${2,1}^2=2,1\cdot 2,1=4,41$   

${\left(-1,1\right)}^2=\left(-1,1\right)\cdot \left(-1,1\right)=1,21$  

${0,03}^2=0,03\cdot 0,03=0,0009$  

${\left(-1,2\right)}^2=\left(-1,2\right)\cdot \left(-1,2\right)=1,44$  

${0,1}^2=0,1\cdot 0,1=0,01$