Question

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Answer 1

31.2

2) (8+у) * (у-8) = (у+8) * (у-8) = у² - 64

4) $(a+\frac{2}{3}b) * (a-\frac{2}{3}b) =a^{2} - (\frac{2}{3}b)^{2} = a^{2}- \frac{4}{9} b^{2}$

6) $(\frac{4}{15} n-m) * (m+\frac{4}{15} n)=(\frac{4}{15} n-m) * (\frac{4}{15} n+m) = \frac{16}{225} n^{2} -m^{2}$

8) (-4a+3b) * (3b+4a) = (3b - 4a) * (3b +4a) = 9b² - 16a²

10) $(\frac{5}{4} c + \frac{3}{7}d) * (\frac{3}{7}d-\frac{5}{4}c) = (\frac{3}{7} d+\frac{5}{4}c) * (\frac{3}{7}d-\frac{5}{4}c) = \frac{9}{49}d^{2} -\frac{25}{16}c^{2}$

12) $(\frac{1}{5} a+\frac{1}{9} b)*(\frac{1}{9} b - \frac{1}{5} a) = (\frac{1}{9}b+\frac{1}{5}a)*(\frac{1}{9} b-\frac{1}{5} a)= \frac{1}{81}b^{2} - \frac{1}{25}a^{2}$

31.3

2) 64-b² = 8² - b² = (8-b) * (8+b)

4) 2,89 - d² = $\frac{289}{100}$-d² = $\frac{1}{100} * (289-100d^{2}) = \frac{1}{100} * (17-10d) * (17+10d)$

6) $\frac{100}{121} -y^{2}=\frac{1}{121}8(100-121y^{2}) = \frac{1}{121} * (10-11y) * (10+11y)$

8) $t^{2} -\frac{400}{441} = \frac{1}{441} 8(441t^{2} - 400) = \frac{1}{441}* (21t-20) * (21t+20)$

10) -16 + 49y² = 49y - 16 = (7y-4) * (7y+4)

12) $\frac{4}{25}t^{2}-36=\frac{4}{25}*(t^{2}-25*9)=\frac{4}{25}*(t^{2} - 225) = \frac{4}{25}* (t-15) * (t+15)$