Zad 12 str 24 Matematyka z plusem 3 zad 12: Dla każdej z podanych liczb znajdź dwie kolejne liczby całkowite, tak aby jedna z nich była mniejsza, a druga większa od tej liczby.

[latex]sqrt{11}; sqrt{23}; 2sqrt{50}; sqrt[3]{9}; -sqrt{15}; sqrt{0,6}; -sqrt[3]{30}[/latex] ========== [latex]sqrt{9} extless sqrt{11} extless sqrt{16} \\ 3 extless sqrt{11} extless 4[/latex] ---------- [latex]sqrt{16} extless sqrt{23} extless sqrt{25} \\ 4 extless sqrt{23} extless 5[/latex] ---------- [latex]2sqrt{50} = sqrt{4} cdot sqrt{50} = sqrt{ 4 cdot 50}=sqrt{200}\\ sqrt{196} extless sqrt{200} extless sqrt{225} \\ 14 extless sqrt{200} extless 15 \\ 14 extless 2sqrt{50} extless 15[/latex] ---------- [latex]sqrt[3]{8} extless sqrt[3]{9} extless sqrt[3]{27} \\ 2 extless sqrt[3]{9} extless 3[/latex] ---------- [latex]sqrt{9} extless sqrt{15} extless sqrt{16} \\ 3 extless sqrt{15} extless 4 \\ -3 extgreater - sqrt{15} extgreater - 4 \\ -4 extless -sqrt{15} extless -3[/latex] ---------- [latex]sqrt{0} extless sqrt{0,6} extless sqrt{1} \\ 0 extless sqrt{0,6} extless 1[/latex] ---------- [latex]sqrt[3]{27} extless sqrt[3]{30} extless sqrt[3]{64} \\ 3 extless sqrt[3]{30} extless 4 \\ -3 extgreater -sqrt[3]{30} extgreater - 4 \\ -4 extless -sqrt[3]{30} extless -3[/latex]