Odpowiedź:
11) 3(√5)² = 3*5 = 15
12) (-6√2)² = 36*2=72
13) 6√5*√5 = 6(√5)² = 6*5 = 30
14) 3√2*5√2 = 3*5*(√2)² = 15*2 = 30
15) (3/7√14)² = 9/49*14 = 9*2/7 = 18/7 = 2 4/7
16) (-0,6√30)² = 0,36*30 = 10,8
17) -2√5(6√5-3) = -12(√5)² + 6√5 = -12*5 + 6√5 = -60 + 6√5 ≈ -46,583592135
18) (2√5-3)(4+3√5) = 8√5 + 6(√5)² - 12 - 9√5 = 8√5 + 6*5 - 12 - 9√5 = 8√5 + 30 - 12 - 9√5 = 18 - √5
19) -7(3√2+1) - (4-√2)(4√2+1) = -21√2 - 7 - (16√2+4-4(√2)²-√2 =
-21√2 - 7 - 16√2 - 4 +4*2 + √2 = -36√2 - 3 = -3-36√2 ≈ -53.9116882454
20) (-√3)^6 = (-√3)(-√3)(-√3)(-√3)(-√3)(-√3) = (-√3)² * (-√3)² * (-√3)² = 3*3*3 = 27
21) ∛12^6 = ∛2985984 = 144 lub
∛12^6 = ∛(12²)³ = 12² = 144
-----------
Szczegółowe wyjaśnienie:
[tex]11) \ 3(\sqrt{5})^{2} = 3\cdot5 = 15\\\\12) \ (-6\sqrt{2})^{2} = (-6)^{2}\cdot(\sqrt{2})^{2} = 36\cdot2 =72\\\\13) \ 6\sqrt{5}\cdot\sqrt{5} = 6\sqrt{5\cdot5} = 6\sqrt{5^{2}} = 6\cdot5 = 30\\\\14) \ 3\sqrt{2}\cdot5\sqrt{2} = 15\sqrt{2\cdot2} = 15\sqrt{2^{2}}=15\cdot2 = 30\\\\15) \ (\frac{3}{7}\sqrt{14})^{2} = (\frac{3}{7})^{2}\cdot(\sqrt{14})^{2} = \frac{9}{49}\cdot14 = \frac{9}{7}\cdot2 = \frac{18}{7} = 2\frac{4}{7}[/tex]
[tex]16) \ (-0,6\sqrt{30})^{2} =(-0,6)^{2}\cdot(\sqrt{30})^{2} = 0,36\cdot30 = 10,8[/tex]
[tex]17) \ -2\sqrt{5}(6\sqrt{5}-3) = -2\sqrt{5}\cdot6\sqrt{5}-2\sqrt{5}\cdot(-3) = -12\sqrt{5\cdot5}+6\sqrt{5}=\\\\=-12\cdot5+6\sqrt{5}=6\sqrt{5}-60=6(\sqrt{5}-10)[/tex]
[tex]18) \ (2\sqrt{5}-3)(4+3\sqrt{5}) = 2\sqrt{5}\cdot4+2\sqrt{5}\cdot3\sqrt{5}-3\cdot4-3\cdot3\sqrt{5} =\\\\=8\sqrt{5}+6\sqrt{5\cdot5}-12-9\sqrt{5} =8\sqrt{5}+6\cdot5-12-9\sqrt{5} =8\sqrt{5}+30-12-9\sqrt{5}=\\\\=18-\sqrt{5}[/tex]
[tex]19) \ -7(3\sqrt{2}+1)-(4-\sqrt{2})(4\sqrt{2}+1) =\\\\= -7\cdot3\sqrt{2}-7\cdot1-(4\cdot4\sqrt{2}+4\cdot1-\sqrt{2}\cdot4\sqrt{2}-\sqrt{2}\cdot1)\\\\=-21\sqrt{2}-7-(16\sqrt{2}+4-4\sqrt{2\cdot2}-\sqrt{2})=-21\sqrt{2}-7-(15\sqrt{2}+4-4\cdot2)=\\\\=-21\sqrt{2}-7-(15\sqrt{2}-4) = -21\sqrt{2}-7-15\sqrt{2}+4 = -3-36\sqrt{2}[/tex]
[tex]20) \ (-\sqrt{3})^{6} = (\sqrt{3})^{6} = (3^{\frac{1}{2}})^{6}=3^{\frac{1}{2}\cdot6} = 3^{3} = 27\\\\21) \ \sqrt[3]{12^{6}} = 12^{\frac{6}{3}} = 12^{2} = 144[/tex]
Wyjasnienie
Korzystamy z własności pierwiastków:
[tex](\sqrt[n]{a})^{n} = a\\\\\sqrt[n]{a}\cdot\sqrt[n]{b} = \sqrt[n]{a\cdot b}[/tex]
Ponadto: liczba ujemna podniesiona do potęgi parzystej daje liczbę dodatnią.
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Odpowiedź:
11) 3(√5)² = 3*5 = 15
12) (-6√2)² = 36*2=72
13) 6√5*√5 = 6(√5)² = 6*5 = 30
14) 3√2*5√2 = 3*5*(√2)² = 15*2 = 30
15) (3/7√14)² = 9/49*14 = 9*2/7 = 18/7 = 2 4/7
16) (-0,6√30)² = 0,36*30 = 10,8
17) -2√5(6√5-3) = -12(√5)² + 6√5 = -12*5 + 6√5 = -60 + 6√5 ≈ -46,583592135
18) (2√5-3)(4+3√5) = 8√5 + 6(√5)² - 12 - 9√5 = 8√5 + 6*5 - 12 - 9√5 = 8√5 + 30 - 12 - 9√5 = 18 - √5
19) -7(3√2+1) - (4-√2)(4√2+1) = -21√2 - 7 - (16√2+4-4(√2)²-√2 =
-21√2 - 7 - 16√2 - 4 +4*2 + √2 = -36√2 - 3 = -3-36√2 ≈ -53.9116882454
20) (-√3)^6 = (-√3)(-√3)(-√3)(-√3)(-√3)(-√3) = (-√3)² * (-√3)² * (-√3)² = 3*3*3 = 27
21) ∛12^6 = ∛2985984 = 144 lub
∛12^6 = ∛(12²)³ = 12² = 144
-----------
Szczegółowe wyjaśnienie:
[tex]11) \ 3(\sqrt{5})^{2} = 3\cdot5 = 15\\\\12) \ (-6\sqrt{2})^{2} = (-6)^{2}\cdot(\sqrt{2})^{2} = 36\cdot2 =72\\\\13) \ 6\sqrt{5}\cdot\sqrt{5} = 6\sqrt{5\cdot5} = 6\sqrt{5^{2}} = 6\cdot5 = 30\\\\14) \ 3\sqrt{2}\cdot5\sqrt{2} = 15\sqrt{2\cdot2} = 15\sqrt{2^{2}}=15\cdot2 = 30\\\\15) \ (\frac{3}{7}\sqrt{14})^{2} = (\frac{3}{7})^{2}\cdot(\sqrt{14})^{2} = \frac{9}{49}\cdot14 = \frac{9}{7}\cdot2 = \frac{18}{7} = 2\frac{4}{7}[/tex]
[tex]16) \ (-0,6\sqrt{30})^{2} =(-0,6)^{2}\cdot(\sqrt{30})^{2} = 0,36\cdot30 = 10,8[/tex]
[tex]17) \ -2\sqrt{5}(6\sqrt{5}-3) = -2\sqrt{5}\cdot6\sqrt{5}-2\sqrt{5}\cdot(-3) = -12\sqrt{5\cdot5}+6\sqrt{5}=\\\\=-12\cdot5+6\sqrt{5}=6\sqrt{5}-60=6(\sqrt{5}-10)[/tex]
[tex]18) \ (2\sqrt{5}-3)(4+3\sqrt{5}) = 2\sqrt{5}\cdot4+2\sqrt{5}\cdot3\sqrt{5}-3\cdot4-3\cdot3\sqrt{5} =\\\\=8\sqrt{5}+6\sqrt{5\cdot5}-12-9\sqrt{5} =8\sqrt{5}+6\cdot5-12-9\sqrt{5} =8\sqrt{5}+30-12-9\sqrt{5}=\\\\=18-\sqrt{5}[/tex]
[tex]19) \ -7(3\sqrt{2}+1)-(4-\sqrt{2})(4\sqrt{2}+1) =\\\\= -7\cdot3\sqrt{2}-7\cdot1-(4\cdot4\sqrt{2}+4\cdot1-\sqrt{2}\cdot4\sqrt{2}-\sqrt{2}\cdot1)\\\\=-21\sqrt{2}-7-(16\sqrt{2}+4-4\sqrt{2\cdot2}-\sqrt{2})=-21\sqrt{2}-7-(15\sqrt{2}+4-4\cdot2)=\\\\=-21\sqrt{2}-7-(15\sqrt{2}-4) = -21\sqrt{2}-7-15\sqrt{2}+4 = -3-36\sqrt{2}[/tex]
[tex]20) \ (-\sqrt{3})^{6} = (\sqrt{3})^{6} = (3^{\frac{1}{2}})^{6}=3^{\frac{1}{2}\cdot6} = 3^{3} = 27\\\\21) \ \sqrt[3]{12^{6}} = 12^{\frac{6}{3}} = 12^{2} = 144[/tex]
Wyjasnienie
Korzystamy z własności pierwiastków:
[tex](\sqrt[n]{a})^{n} = a\\\\\sqrt[n]{a}\cdot\sqrt[n]{b} = \sqrt[n]{a\cdot b}[/tex]
Ponadto: liczba ujemna podniesiona do potęgi parzystej daje liczbę dodatnią.