Hej... Proszę o rozwiązanie zadań z załącznika ( 1,2,5,6). Z góry dziękuję. Daję naj.
1) [(125)¹/³·(⅕)⁻²·5¹/²] / (√5)⁻¹= [(5³)¹/³·5²·5¹/²] / (5⁻¹/²)= [5·5²·5¹/²] /(5⁻¹/²) =
= (5³ ¹/²)/(5⁻¹/²) = 5⁴
2) log₆ x = 2log₂ √3 - log₂ 12
log₆ x = log₂ (√3)² - log₂ 12
log₆ x = log₂ 3 - log₂ 12
log₆ x = log₂ ¼
log₆ x = -2 ⇒ x=6⁻² = (⅙)² = 1/36
5) 1/(2+√3) +(2-√3)⁻¹ = 1/(2+√3) + 1/(2-√3) = (sprowadzamy do wspólnego mianownika)
= (2-√3)/[(2+√3)(2-√3)] + (2+√3)/[(2-√3)(2+√3)] =
= (2-√3+2+√3)/ (4-3) = 4/1 = 4 ∈W
6) 5-2√3 = √(37-20√3) (podnosimy obustronnie do kwadratu)
(5-2√3)² = 37-20√3
L = 25-20√3+(2√3)² = 25-20√3+12 = 37-20√3
P = 37-20√3
L = P
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1) [(125)¹/³·(⅕)⁻²·5¹/²] / (√5)⁻¹= [(5³)¹/³·5²·5¹/²] / (5⁻¹/²)= [5·5²·5¹/²] /(5⁻¹/²) =
= (5³ ¹/²)/(5⁻¹/²) = 5⁴
2) log₆ x = 2log₂ √3 - log₂ 12
log₆ x = log₂ (√3)² - log₂ 12
log₆ x = log₂ 3 - log₂ 12
log₆ x = log₂ ¼
log₆ x = -2 ⇒ x=6⁻² = (⅙)² = 1/36
5) 1/(2+√3) +(2-√3)⁻¹ = 1/(2+√3) + 1/(2-√3) = (sprowadzamy do wspólnego mianownika)
= (2-√3)/[(2+√3)(2-√3)] + (2+√3)/[(2-√3)(2+√3)] =
= (2-√3+2+√3)/ (4-3) = 4/1 = 4 ∈W
6) 5-2√3 = √(37-20√3) (podnosimy obustronnie do kwadratu)
(5-2√3)² = 37-20√3
L = 25-20√3+(2√3)² = 25-20√3+12 = 37-20√3
P = 37-20√3
L = P