Question

va rog sa ma ajutati repedee. doar ce e incercuit

Answer 1

Uiteeee sper ca te am ajutat

Answer 2

f) $\frac{\left(\left(\frac{3}{5}\right)^3\right)^7}{\left(\frac{3}{5}\right)^{20}}+\frac{2}{5}=1$

$\left(\left(\frac{3}{5}\right)^3\right)^7 =\left(\frac{3}{5}\right)^{3\cdot \:7} =\left(\frac{3}{5}\right)^{21} =\frac{3^{21}}{5^{21}}$

$\left(\frac{3}{5}\right)^{20}=\frac{3^{20}}{5^{20}}$

$\frac{\frac{3^{21}}{5^{21}}}{\frac{3^{20}}{5^{20}}} =\frac{3^{21}\cdot \:5^{20}}{5^{21}\cdot \:3^{20}} =\frac{5^{20}\cdot \:3^{21-20}}{5^{21}} =\frac{5^{20}\cdot \:3}{5^{21}} =\frac{3}{5^{21-20}} =\frac{3}{5}$

$\frac{3}{5}+\frac{2}{5} =\frac{3+2}{5} =\frac{5}{5} =1$

g) $\left(\frac{4}{3}\right)^2-\left(\frac{7}{6}\right)^2 =\frac{4^2}{3^2}-\left(\frac{7}{6}\right)^2 =\frac{4^2}{3^2}-\frac{7^2}{6^2} =\frac{4^2}{9}-\frac{7^2}{6^2} =\frac{4^2}{9}-\frac{7^2}{36}=\frac{64}{36}-\frac{7^2}{36} =\frac{64-7^2}{36} =\frac{15}{36} =\frac{5}{12}$

h) $\frac{4+\left(\frac{3}{2}\right)^2}{\left(\frac{5}{2}\right)^2} =\frac{4+\left(\frac{3}{2}\right)^2}{\frac{5^2}{2^2}} =\frac{4+\frac{3^2}{2^2}}{\frac{5^2}{2^2}} =\frac{4+\frac{9}{4}}{\frac{5^2}{2^2}} =\frac{\left(4+\frac{9}{4}\right)\cdot \:2^2}{5^2}=\frac{2^2\cdot \frac{25}{4}}{5^2} =\frac{4\cdot \frac{25}{4}}{5^2} =\frac{4\cdot \frac{25}{4}}{25} =\frac{25}{25}=1$

4. a)  $3-\left(\frac{6}{5}\right)^2 =3-\frac{6^2}{5^2} =3-\frac{36}{25} =\frac{3\cdot \:25}{25}-\frac{36}{25} =\frac{3\cdot \:25-36}{25} =\frac{39}{25}$

b)$5-\left(\frac{21}{10}\right)^2 =5-\frac{21^2}{10^2} =5-\frac{441}{100} =\frac{5\cdot \:100-441}{100}=\frac{59}{100}$

c) $4-\left(\left(\frac{5}{2}\right)^2-\frac{23}{10}^2\right)=$

$\left(\frac{5}{2}\right)^2-\left(\frac{23}{10}\right)^2 =\frac{5^2}{2^2}-\left(\frac{23}{10}\right)^2 =\frac{5^2}{4}-\frac{23^2}{10^2} =\frac{5^2}{4}-\frac{23^2}{100} =\frac{625}{100}-\frac{23^2}{100} =\frac{625-23^2}{100}=\frac{96}{100}=\frac{24}{25}$

$=4-\frac{24}{25} =\frac{4\cdot \:25}{25}-\frac{24}{25} =\frac{4\cdot \:25-24}{25} =\frac{76}{25}$

Sper că te-am ajutat!

Succes!

#copăceibrainly