Rozwiązanie
[tex]\sqrt{49}=7, \ \text{bo} \ 7^2=7\cdot7=49\\\\\sqrt{400}=20, \ \text{bo} \ 20^2=20\cdot20=400\\\\\sqrt{3\frac{1}{16}}=\sqrt{\frac{49}{16}}=\frac{7}{4}=1\frac{3}{4}, \ \text{bo} \ (1\frac{3}{4})^2=(\frac{7}{4})^2=\frac{49}{16}=3\frac{1}{16}\\\\\sqrt{0,0036}=0,06, \ \text{bo} \ 0,06^2=0,06\cdot0,06=0,0036\\\\\sqrt0=0, \ \text{bo} \ 0^2=0\cdot0=0\\\\\sqrt{0,16}=0,4, \ \text{bo} \ 0,4^2=0,4\cdot0,4=0,16\\\\\sqrt{1,69}=1,3, \ \text{bo} \ 1,3^2=1,3\cdot1,3=1,69[/tex]
