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Решите задачу:

$(-2\frac{4}{9}a^2b+2\frac{1}{5}a^3b^2-ab^5)(-2\frac{1}{22}a^4b^5)=\\\\=(-\frac{22}{9}a^2b+\frac{11}{5}a^3b^2-ab^5)*(-\frac{45}{22}a^4b^5)=\\\\=-\frac{22}{9}a^2b*(-\frac{45}{22}a^4b^5)+\frac{11}{5}a^3b^2*(-\frac{45}{22}a^4b^5)-ab^5*(-\frac{45}{22}a^4b^5)=\\\\\frac{22*5*9}{9*22}a^{2+4}b^{1+5}-\frac{11*5*9}{11*2*5}a^{3+4}b^{2+5}+\frac{45}{22}a^{1+4}b^{5+5}=\\\\5a^6b^6-4.5a^7b^7+\frac{45}{22}a^5b^{10}$