Решить √50-√32-1/3√18. (√2+3)²-3√8. (2-√3)²+2√12. (3√10-12/√10):(2√10-19/√10).. Created by ifeoktistov3. matematika-ru

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July 2022 1 25 Report

Решить √50-√32-1/3√18.
(√2+3)²-3√8.
(2-√3)²+2√12.
(3√10-12/√10):(2√10-19/√10).

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93931 √50 - √32 - ¹/₃√18 = √50 - √32 - √¹/₉*18 =√50 - √32 - √2 = √25*2 - √16*2 - √2 = 5√2 - 4√2 - √2 = 0√2 = 0.
(√2+3)²-3√8 = 2 + 2*√2*3 + 9 - 3√8 = 11 + 6√2 - 3√8 = 11 + 6√2 - 3√4*2 =
11 + 6√2 - 3*2√2 = 11 + 6√2 - 6√2 = 11 + 0 = 11.
(2-√3)²+2√12 = 4 - 2*2*√3 + 3 + 2√12 = 7 - 4√3 + 2√12 = 7 - 4√3 + 2√4*3 =
7 - 4√3 + 4√3 = 7 + 0 = 7.
(3√10 - ¹²/√10) : (2√10 - ¹⁹/√10) = 3√10 : 2√10 - ¹²/√10 : (-¹⁹/√10) =
³/₂ - ¹²/√10 * (- √10/₁₉) = 1¹/₂ + ¹²/₁₉ = 1¹⁹/₃₈ + ²⁴/₃₈ = 1⁴³/₃₈ = 2⁵/₃₈.

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