Question

помогите решить пж пж помогите даю 90 б​

Answer 1

Объяснение:

5.55

$1)\ 2,1*1,9=(2+0,1)*(2-0,1)=2^2-0,1^2=4-0,01=3,99.\\2)\ 4,02*3,98=(4+0,02)*(4-0,02)=4^2-0,02^2=16-0,0004=15,9996.\\3)\ 19,8*20,2=(20-0,2)*(20+0,2)=20^2-0,2^2=400-0,04=399,96.\\4)\ 1,05*0,95=(1+0,05)*(1-0,05)=1^2-0,05^2=1-0,0025=0,9975.\\5)\ (3\frac{2}{3})^2-(2\frac{1}{3})^2=(3\frac{2}{3}+2\frac{1}{3})*(3\frac{2}{3}-2\frac{1}{3})=6*1\frac{1}{3}=6*\frac{4}{3}=8.$

$6)\ (4\frac{1}{6})^2-(1\frac{1}{6})^2=(4\frac{1}{6} -1\frac{1}{6})*(4\frac{1}{6}+1\frac{1}{6})=3*5\frac{1}{3}=3*\frac{16}{3}=16.\\7)\ (5\frac{2}{3})^2-(4\frac{1}{3})^2=(5\frac{2}{3}+4\frac{1}{3})*(5\frac{2}{3}-4\frac{1}{3})=10*1\frac{1}{3}=\frac{40}{3}=13\frac{1}{3}.\\8)\ 21,3^2-21,2^2=(21,3-21,2)*(21,3+21,2)=0,1*42,5=4,25.$

55.6

$1)\ \frac{72}{13^2-11^2}=\frac{72}{(13-11)*(13+11)}=\frac{72}{2*24}=\frac{3}{2}=1,5.\\2)\ \frac{79^2-65^2}{420}= \frac{(79-65)*(79+65)}{420} =\frac{14*144}{14*30}=4,8.\\ 3)\ \frac{92^2-48^2}{27^2-17^2}=\frac{(92-48)*(92+48)}{(27-17)*(27+17)}=\frac{44*140}{10*44} =14.\\4)\ \frac{63^2-27^2}{83^2-79^2}=\frac{(63-27)*(63+27)}{(83-79)*(83+79)}=\frac{36*90}{4*162}=5.$