5,6,7,8 нужен ответы

$5)(3 \frac{2}{3})^2- (2\frac{1}{3})^2=(3 \frac{2}{3}-2 \frac{1}{3})(3 \frac{2}{3}+2 \frac{1}{3})=1 \frac{1}{3}*6= \frac{4*6}{3}= \frac{24}{3}=8;$
$6)(4 \frac{1}{6})^2-(1 \frac{1}{6})^2=(4 \frac{1}{6}-1 \frac{1}{6})( 4\frac{1}{6}+1 \frac{1}{6})=3*5 \frac{1}{3}= \frac{3*16}{3}=16;$
$7)(5 \frac{2}{3})^2-(4 \frac{1}{3})^2=(5 \frac{2}{3}-4 \frac{1}{3})(5 \frac{2}{3}+4 \frac{1}{3})=1 \frac{1}{3}*10= \frac{4*10}{3}= \frac{40}{3}=13 \frac{1}{3};$
$8)21,3^2-21,2^2=(21,3-21,2)(21,3+21,2)=0,1*42,5=4,25;$
Применял формулу: $a^2-b^2=(a-b)(a+b);$