Help please 8cos(-390°)-sin390° * tg1035°

Question

Алгебра: Help please 8cos(-390°)-sin390° * tg1035°

MoskowCorp · Accepted Answer

Инайдите числовое значение выражения: a) x+y)²y /(x-y)(x+y)(x²+y²) +x/(x²+y²) при x =0,3 ; y =1/2 (x+y)²y /(x-y)(x+y)(x²+y²) +x/(x²+y²) =(x+y)y /(x-y)(x²+y²) +x/ (x²+y²) =(xy+y² +x² -xy)/(x²+y²)(x-y) =(y² +x²)/ (x²+y²)(x-y) =1/(x-y) =1/(0,3 -0,5) = 1/(-0,2) = -5. б) (x³+y³) /(x² -y²) -(x²+y²) / (x-y) при x =0,4 ; y =1/2 (x³+y³) /(x² -y²) -(x²+y²) / (x-y) =(x+y)(x² -xy +y²) /(x-y)(x+y) - (x²+y²) / (x-y) =(x² -xy +y²) /(x-y) -(x²+y²) / (x-y) = (x² -xy +y² -( x²+y²)) /(x-y) = - xy/(x-y) == -0,4*0,5...

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