Question

Va rog mult explicati mi si mie dau simulare si eu ma blokez la genul asta de exercitiu!​

Answer 1

$a = \frac{ \sqrt{2} \times \sqrt{3} \times 3}{ \sqrt{2} } + \frac{ \sqrt{2} \times \sqrt{3} \times 5}{ \sqrt{3} } - 5 \sqrt{2} + 7 = \\ = 3 \sqrt{3} + 5 \sqrt{2} - 5 \sqrt{2} + 7 = \\ 3 \sqrt{3} + 7$
b=3(2+√3)+2=8+3√3
(a-b)^2018=3√3+7-8-3√3=(-1)^2018=1

Answer 2

$notam \: a \: = \: x \: si \: b \: = \: y \\ \\ x = \sqrt{6} \times ( \frac{3}{ \sqrt{2} } + \frac{5}{ \sqrt{3} } ) - |5 \sqrt{2} - 7 | \\ prima \: oara \: rationalizam \\ x = \sqrt{6} \times ( \frac{3 \sqrt{2} }{2} + \frac{5 \sqrt{3} }{3} ) - |5 \sqrt{2} - 7| \\ apoi \: aducem \: la \: acelasi \: numitor \: fractiile \: din \: paranteza \\ x = \sqrt{6} \times \frac{9 \sqrt{2} + 10 \sqrt{3} }{6} - |5 \sqrt{2} - 7| \\ stabilim \: care \: numar \: e \: mai \: mare \: din \: modul \\ 5 \sqrt{2} = \sqrt{5 {}^{2} \times 2 } = \sqrt{25 \times 2} = \sqrt{50} \\ 7 = \sqrt{7 {}^{2} } = \sqrt{49} \\ = > \sqrt{50} > \sqrt{49} = > |5 \sqrt{2} - 7| = 5 \sqrt{2} - 7 \\ x = \sqrt{6} \times \frac{9 \sqrt{2} + 10 \sqrt{3} }{6} - (5 \sqrt{2} - 7) \\ x = \frac{ \sqrt{6} \times (9 \sqrt{2} + 10 \sqrt{3} )}{6} - 5 \sqrt{2} + 7 \\ x = \frac{9 \sqrt{12} + 10 \sqrt{18} }{6} - 5 \sqrt{2} + 7 \\ x = \frac{18 \sqrt{3} + 30 \sqrt{2} }{6} - 5 \sqrt{2} + 7 \\ dam \: factor \: comun \\ x = \frac{6(3 \sqrt{3} + 5 \sqrt{2}) }{6} - 5 \sqrt{2} + 7 = > x = 3 \sqrt{3} + 5 \sqrt{2} - 5 \sqrt{2} + 7 \\ x = 3 \sqrt{3} + 7$

$y = \frac{3}{2 - \sqrt{3} } + ( \sqrt{2} ) {}^{2} \\ rationalizam \: cu \: conjugata \\ y = \frac{3(2 + \sqrt{3}) }{4 - 3} + 2 \\ y = \frac{6 + 3 \sqrt{3} }{1} + 2 \\ y = 3 \sqrt{3} + 8 \\ \\ (x - y) {}^{2018} = (3 \sqrt{3} + 7 - (3 \sqrt{3} + 8)) {}^{2018} = \\ = (3 \sqrt{3} + 7 - 3 \sqrt{ 3} - 8) {}^{2018} = ( - 1) {}^{2018} = 1$