Question:`x^2 - 2sqrt(42) - 13 = 0; x > 0` ক. দেখাও যে, `x = sqrt(7) + sqrt(6)` খ. প্রমাণ কর যে, `x^3 + 1/x^3 = 50sqrt(7)` গ. `(x^5 - 1/x^5)` এর মান নির্ণয় কর।
Answer ক. দেওয়া আছে, `x^2 - 2sqrt(42) - 13 = 0` বা, `x^2 = 13 + 2sqrt(42)` বা, `x^2 = 7 + 2sqrt(42) + 6` বা, `x^2 = (sqrt(7))^2 + 2.sqrt(7).sqrt(6) + (sqrt(6))^2` বা, `x^2 = (sqrt(7) + sqrt(6))^2` `:. x = sqrt(7) + sqrt(6)` (দেখানো হলো) খ. এখন, `1/x = 1/(sqrt(7) + sqrt(6)` বা, `1/x = (1. sqrt(7) - sqrt(6))/((sqrt(7) + sqrt(6)) (sqrt(7) - sqrt(6))` বা, `1/x = (sqrt(7) - sqrt(6))/((sqrt(7))^2 - (sqrt(6))^2` বা, `1/x = (sqrt(7) - sqrt(6))/(7 - 6)` বা, `1/x = sqrt(7) - sqrt(6)` `:. x + 1/x = sqrt(7) + sqrt(6) + sqrt(7) - sqrt(6)` `:. x + 1/x = 2sqrt(7)` বামপক্ষ `= x^3 + 1/x^3` `= (x + 1/x)^3 - 3.2sqrt(7)` [মান বসিয়ে] `= 56sqrt(7) - 6sqrt(7)` `= 50sqrt(7)` = ডানপক্ষ `:. x^3 + 1/x^3 = 50sqrt(7)` গ. এখন, `x - 1/x = (sqrt(7) + sqrt(6)) - (sqrt(7) - sqrt(6))` `= sqrt(7) + sqrt(6) - sqrt(7) + sqrt(6) = 2sqrt(6)` `:. x^2 + 1/x^2 = (x + 1/x)^2 - 2.x.1/x` `= (2sqrt(7))^2 - 2` [মান বসিয়ে] `= 28 - 2 = 26` এখন, `x^3 - 1/x^3 = (x - 1/x)^3 + 3.x.1/x (x - 1/x)` `= (2sqrt(6))^3 + 3.2sqrt(6)` [মান বসিয়ে] ` = 48sqrt(6) + 6sqrt(6) = 54sqrt(6)` `:. (x^2 + 1/x^2) (x^3 - 1/x^3) = 26 xx 54sqrt(6)` বা, `x^5 - 1/x + x - 1/x^3 = 1404sqrt(6)` বা, `(x^5 + 1/x^5) + (x - 1/x) = 1404sqrt(6)` বা, `x^5 - 1/x^5 + 2sqrt(6) = 1404sqrt(6)` বা, `x^5 - 1/x^5 + 2sqrt(6) = 1404sqrt(6)` বা, `x^5 - 1/x^5 = 1404sqrt(6) - 2sqrt(6) = 1402sqrt(6)` `:. x^5 - 1/x^5 = 1402sqrt(6)` (Ans)
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`x^2 - 2sqrt(42) - 13 = 0; x > 0` ka. dekhao je, `x = sqrt(7) + sqrt(6)` kh. proman kar je, `x^3 + 1/x^3 = 50sqrt(7)` ga. `(x^5 - 1/x^5)` ar man nirony karo.