Question

calculati dau 98 puncte​

Answer 1

c) 4/3×{2/3-[4/5-(1/2+2/3)×2/7]×5/14}=

=4/3×{2/3-[4/5-(3/6+4/6)×2/7]×5/14}=

=4/3×[2/3-(4/5-7/6×2/7)×5/14]=

=4/3×[2/3-(4/5-1/3)×5/14]=

=4/3×[2/3-(12/15-5/15)×5/14]=

=4/3×(2/3-7/15×5/14)=

=4/3×(2/3-1/6)=

=4/3×(4/6-1/6)=

=4/3×3/6=2/3

Answer 2

$\bf \dfrac{4}{3}\cdot \Big[ \dfrac{2}{3}-\Big( \dfrac{4}{5}-\Big(\dfrac{1}{2}-\dfrac{2}{3}\Big)\cdot\dfrac{2}{7}\Big)\cdot\dfrac{5}{14}\Big]=$

$\bf \dfrac{4}{3}\cdot \Big[ \dfrac{2}{3}-\Big( \dfrac{4}{5}-\dfrac{7}{6}\cdot\dfrac{2}{7}\Big)\cdot\dfrac{5}{14}\Big]=$

$\bf \dfrac{4}{3}\cdot \Big[ \dfrac{2}{3}-\Big( \dfrac{4}{5}-\dfrac{\not7}{\not6}\cdot\dfrac{\not2}{\not7}\Big)\cdot\dfrac{5}{14}\Big]=$

$\bf \dfrac{4}{3}\cdot \Big[ \dfrac{2}{3}-\Big( \dfrac{4}{5}-\dfrac{1}{3}\Big)\cdot\dfrac{5}{14}\Big]=$

$\bf \dfrac{4}{3}\cdot \Big[ \dfrac{2}{3}-\Big( \dfrac{4\cdot3}{5\cdot3}-\dfrac{1\cdot5}{3\cdot5}\Big)\cdot\dfrac{5}{14}\Big]=$

$\bf \dfrac{4}{3}\cdot \Big( \dfrac{2}{3}- \dfrac{12-5}{15}\cdot\dfrac{5}{14}\Big)=$

$\bf \dfrac{4}{3}\cdot \Big( \dfrac{2}{3}- \dfrac{7}{15}\cdot\dfrac{5}{14}\Big)=$

$\bf \dfrac{4}{3}\cdot \Big( \dfrac{2}{3}- \dfrac{\not7}{\not15}\cdot\dfrac{\not5}{\not14}\Big)=$

$\bf \dfrac{4}{3}\cdot \Big( \dfrac{2}{3}- \dfrac{1}{3}\cdot\dfrac{1}{2}\Big)=$

$\bf \dfrac{4}{3}\cdot \Big( \dfrac{2}{3}- \dfrac{1}{6}\Big)=$

$\bf \dfrac{4}{3}\cdot \Big( \dfrac{2\cdot2}{3\cdot2}- \dfrac{1}{6}\Big)=$

$\bf \dfrac{4}{3}\cdot \dfrac{4-1}{6}=$

$\bf \dfrac{4}{3}\cdot \dfrac{3}{6}=$

$\bf \dfrac{\not4}{\not3}\cdot \dfrac{\not3}{\not6}=$

$\bf \dfrac{2}{1}\cdot \dfrac{1}{3}=$

$\boxed{\bf \dfrac{2}{3}}$