Odpowiedź:
[tex]0,6b(3\frac{1}{3}a^2-7ab)-0,4ab(2\frac{1}{2}a+2b)=\frac{3}{5}b(\frac{10}{3}a^2-7ab)-\frac{2}{5}ab(\frac{5}{2}a+2b)=\\\\=\frac{\not3^1}{\not5_{1}}b\cdot\frac{\not10^2}{\not3_{1}}a^2-\frac{3}{5}b\cdot7ab-\frac{\not2^1}{\not5_{1} }ab\cdot\frac{\not5^1}{\not2_{1}}a-\frac{2}{5}ab\cdot2b=2a^2b-\frac{21}{5}ab^2-a^2b-\frac{4}{5}ab^2=\\\\=a^2b-\frac{25}{5}ab^2=a^2b-5ab^2\\\\Odp.B[/tex]
