Sin 13 пи деленное на 3 + cos 43 пи деленное на 6

Алгебра

Ответить

Ответы

baltgold-m27

18.01.2021

$1)\frac{4a-k}{33k}+\frac{k-3a}{22k}=\frac{8a-2k+3k-9a}{66k}=\frac{k-a}{66k}: \boxed{\frac{k-a}{66k}}$

$2)\frac{2a^{2} }{ab-3b^{2}}-\frac{6a}{a-3b}=\frac{2a^{2}}{b(a-3b)}-\frac{6a}{a-3b}=\frac{2a^{2}-6ab}{b(a-3b)}=\frac{2a(a-3b)}{b(a-3b)}=\frac{2a}{b}: \boxed{\frac{2a}{b}}$

$3)\frac{x-9y}{x^{2}-9y^{2}}-\frac{3y}{3xy-x^{2}}=\frac{x-9y}{(x-3y)(x+3y)}-\frac{3y}{x(3y-x)}=\frac{x-9y}{(x-3y)(x+3y)}+\frac{3y}{x(x-3y)}=\frac{x^{2}-9xy+3xy+9y^{2}}{x(x-3y)(x+3y)}=\frac{x^{2}-6xy+9y^{2}}{x(x-3y)(x+3y)} =\frac{(x-3y)^{2}}{x(x-3y)(x+3y)}=\frac{x-3y}{x(x+3y)}: \boxed{\frac{x-3y}{x(x+3y)}}$

Lenok33lenok89

18.01.2021

Log9 (6√6-15)^2 + log27(6√6+15)^3 = = log3^2 (6√6-15)^2 + log3^3 (6√6+15)^3 = = 1/2* log3 (6√6-15)^2 + 1/3* log3 (6√6+15)^3 = = log3 ((6√6 - 15)^2)^1/2 + log3 ((6√6+15)^3)^1/3 = = log3 | 6√6 - 15 |+ log3 (6√6 +15) = появление модуля = log3 (15 - 6√6) + log3 (15 + 6√6) = = log3 ((15 - 6√6)* (15 + 6√6) ) = = log3 (15^2 - (6√6)^2) = = log3 (225 - 216) = = log3 (9) = = 2