Ответ:
а)
[tex] \sqrt{121} = \sqrt{11 ^{2} } = 11[/tex]
б)
[tex] \sqrt{ \frac{9}{16} } = \sqrt{ \frac{ {3}^{2} }{ {4}^{2} } } = \frac{3}{4} [/tex]
в)
[tex] \sqrt{ \frac{25}{31} } = \frac{5}{ \sqrt{31} } = \frac{5}{ \sqrt{31} } \times \frac{ \sqrt{31} }{ \sqrt{31} } = \frac{5 \sqrt{31} }{31} [/tex]
у̶ м̶е̶н̶я̶ с̶и̶л̶ь̶н̶е̶й̶ш̶е̶е̶ ж̶е̶л̶а̶н̶и̶е̶ н̶а̶п̶и̶с̶а̶т̶ь̶ р̶е̶ш̶е̶н̶и̶е̶ к̶о̶ в̶с̶е̶м̶ з̶а̶д̶а̶н̶и̶я̶м̶, т̶а̶к̶ ч̶т̶о̶ д̶е̶р̶ж̶и̶ е̶щ̶ё̶ е̶х̶е̶х̶е̶
[tex] - 3 \sqrt{2} = - \sqrt{9} \times \sqrt{2} = - \sqrt{18} [/tex]
[tex]7 \sqrt{5} = \sqrt{49} \times \sqrt{5} = \sqrt{245} [/tex]
[tex]2 \sqrt{b} + 6 \sqrt{b} - 11 \sqrt{b} = (2 + 6 - 11) \sqrt{b} = (8 - 11) \sqrt{b} = - 3 \sqrt{b} [/tex]
[tex] \sqrt{36a} + \sqrt{49b} + \sqrt{64a} - \sqrt{81b} = \sqrt{ {6}^{2} a} + \sqrt{ {7}^{2} b} + \sqrt{ {8}^{2} a} - \sqrt{ {9}^{2} b}= 6 \sqrt{a} + 7 \sqrt{b} + 8 \sqrt{a} - 9 \sqrt{b} = 14 \sqrt{a} - 2 \sqrt{b} [/tex]
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Answers & Comments
Ответ:
№1
а)
[tex] \sqrt{121} = \sqrt{11 ^{2} } = 11[/tex]
б)
[tex] \sqrt{ \frac{9}{16} } = \sqrt{ \frac{ {3}^{2} }{ {4}^{2} } } = \frac{3}{4} [/tex]
в)
[tex] \sqrt{ \frac{25}{31} } = \frac{5}{ \sqrt{31} } = \frac{5}{ \sqrt{31} } \times \frac{ \sqrt{31} }{ \sqrt{31} } = \frac{5 \sqrt{31} }{31} [/tex]
у̶ м̶е̶н̶я̶ с̶и̶л̶ь̶н̶е̶й̶ш̶е̶е̶ ж̶е̶л̶а̶н̶и̶е̶ н̶а̶п̶и̶с̶а̶т̶ь̶ р̶е̶ш̶е̶н̶и̶е̶ к̶о̶ в̶с̶е̶м̶ з̶а̶д̶а̶н̶и̶я̶м̶, т̶а̶к̶ ч̶т̶о̶ д̶е̶р̶ж̶и̶ е̶щ̶ё̶ е̶х̶е̶х̶е̶
№2
а)
[tex] - 3 \sqrt{2} = - \sqrt{9} \times \sqrt{2} = - \sqrt{18} [/tex]
б)
[tex]7 \sqrt{5} = \sqrt{49} \times \sqrt{5} = \sqrt{245} [/tex]
№3
а)
[tex]2 \sqrt{b} + 6 \sqrt{b} - 11 \sqrt{b} = (2 + 6 - 11) \sqrt{b} = (8 - 11) \sqrt{b} = - 3 \sqrt{b} [/tex]
б)
[tex] \sqrt{36a} + \sqrt{49b} + \sqrt{64a} - \sqrt{81b} = \sqrt{ {6}^{2} a} + \sqrt{ {7}^{2} b} + \sqrt{ {8}^{2} a} - \sqrt{ {9}^{2} b}= 6 \sqrt{a} + 7 \sqrt{b} + 8 \sqrt{a} - 9 \sqrt{b} = 14 \sqrt{a} - 2 \sqrt{b} [/tex]