Задание 299

Решение:

• $${\largeа)}\ (0{,}1-x)(x+0{,}1)=0{,}1x-x^2+0{,}01-0{,}1x={-}x^2+0{,}01$$

• $${\largeб)}\ (1{,}2-a)(1{,}2+a)=1{,}44-1{,}2a+1{,}2a-a^2=1{,}44-a^2$$

• $${\largeв)}\ \left(\frac{1}{3}-m\right)\left(\frac{1}{2}m-3\right)=\frac{1}{6}m-\frac{1}{2}m^2-1+3m=3\frac{1}{6}m-\frac{1}{2}m^2-1$$

• $${\largeг)}\ \left(\frac{1}{5}a-\frac{3}{7}b\right)(14a+b)=\frac{14}{5}a^2-6ab+\frac{1}{5}ab-\frac{3}{7}b^2=\frac{14}{5}a^2-5\frac{5}{5}ab+\frac{1}{5}ab-\frac{3}{7}b^2=2\frac{4}{5}a^2-5\frac{4}{5}ab-\frac{3}{7}b^2$$

• $${\largeд)}\ (0{,}05y-2{,}3x)(y-0{,}2x)=0{,}05y^2-2{,}3xy-0{,}01xy+0{,}46x^2=0{,}05y^2-2{,}31xy+0{,}46x^2$$

• $${\largeе)}\ (2{,}5a+3b)(0{,}1b-4a)=0{,}25ab+0{,}3b^2-10a^2-12ab=0{,}3b^2-10a^2-11{,}75ab$$

• $${\largeж)}\ \left(\frac{2}{3}m+3n\right)\left(6m-\frac{1}{6}n\right)=4m^2+18mn-\frac{1}{9}mn-\frac{1}{2}n^2=4m^2+17\frac{9}{9}mn-\frac{1}{9}mn-\frac{1}{2}n^2=4m^2+17\frac{8}{9}mn-\frac{1}{2}n^2$$

• $${\largeз)}\ \left(1\frac{1}{2}x-y\right)\left(2\frac{1}{3}y-\frac{1}{3}x\right)=\left(\frac{3}{2}x-y\right)\left(\frac{7}{3}y-\frac{1}{3}x\right)=\frac{7}{2}xy-\frac{7}{3}y^2-\frac{1}{2}x^2+\frac{1}{3}xy=\frac{21}{6}xy-\frac{7}{3}y^2-\frac{1}{2}x^2+\frac{2}{6}xy=\frac{23}{6}xy-\frac{7}{3}y^2-\frac{1}{2}x^2=3\frac{5}{6}xy-2\frac{1}{3}y^2-\frac{1}{2}x^2$$