Question

va implor să mă ajutați​

Answer 1

Răspuns:

$a) {x}^{2} = {3}^{18} + {3}^{21} \\ {x}^{2} = {3}^{18} \times (1 + {3}^{3} ) \\ {x}^{2} = {3}^{18} \times 28 \\ x = \sqrt{ {3}^{18} \times 28 } \\ x = \sqrt{ {3}^{18} } \times \sqrt{28} = \sqrt{ ({3}^{9}) ^{2} } \times \sqrt{4 \times 7} \\ x = {3}^{9} \times 2 \sqrt{7}$

$b) {x}^{2} = {2}^{26} - {2}^{20} \\ {x}^{2} = {2}^{20} \times ( {2}^{6} - 1) \\ {x}^{2} = {2}^{20} \times 63 \\ x = \sqrt{ {2}^{20} \times 63} = \sqrt{ {2}^{20} } \times \sqrt{63} \\ x = \sqrt{ { {(2}^{10} })^{2} } \times \sqrt{7 \times 9} \\ x = {2}^{10} \times 3 \sqrt{7}$

$c) {x}^{2} = {3}^{20} - {3}^{16} \\ {x}^{2} = {3}^{16} \times ( {3}^{4} - 1) \\ {x}^{2} = {3}^{16} \times 80 \\ x = \sqrt{ {3}^{16} \times 80 } = \sqrt{ {3}^{16} } \times \sqrt{80} = \sqrt{ { {(3}^{8} })^{2} } \times \sqrt{16 \times 5} = {3}^{8} \times 4 \sqrt{5}$

$d) {x}^{2} = {5}^{28} - {5}^{26} \\ {x}^{2} = {5}^{26} \times ( {5}^{2} - 1) \\ {x}^{2} = {5}^{26} \times 24 \\ x = \sqrt{ {5}^{26} \times 24} = \sqrt{ {5}^{26} } \times \sqrt{24} = \sqrt{ { {(5}^{13} })^{2} } \times \sqrt{4 \times 6} = {5}^{13} \times 2 \sqrt{6}$

Un exercițiu mai mult de clasa a 5 a, rezultatele nu sunt tocmai plăcute.. Dar de greșit nu am greșit