Question

1 вариант пж очень надо

Answer 1

1.

а) $(-3)^2 - 1^4 = 9 - 1 = \boxed{\bf{8}}$

б) $\left(\dfrac{5}{9}\right)^3 \cdot \left(4\dfrac{1}{2}\right)^2 = \dfrac{5^3}{9^3} \cdot\left(\dfrac{9}{2}\right)^2 = \dfrac{5^3}{9^3}\cdot \dfrac{9^2}{4} = \dfrac{5^3\cdot 9^2}{9^3\cdot 4} = \dfrac{125}{9\cdot 4} = \dfrac{125}{36} = \boxed{\bf{3\dfrac{17}{36}}}$

2.

а) $-27 = \boxed{\bf{(-3)^3}}$

б)  $1,44 = \boxed{\bf{1,2^2}}$

в)  $\dfrac{4}{49} = \boxed{\bf{\left(\dfrac{2}{7}\right)^2}}$

г)  $10000 = \boxed{\bf{10^4}}$

3.

а)  $x^{11}\cdot x^4 = x^{11+4} = \boxed{\bf{x^{15}}}$

б)  $\left(y^3\right)^4= y^{3\cdot 4} = \boxed{\bf{y^{12}}}$

в)  $\dfrac{z^8}{z^4} = z^{8-4} = \boxed{\bf{z^4}}$

г)  $\dfrac{q^{10}\cdot \left(q^3\right)^2}{q^8\cdot q} = \dfrac{q^{10}\cdot q^{3\cdot 2}}{q^{8+1}} = \dfrac{q^{10}\cdot q^6}{q^9} = \dfrac{q^{10+6}}{q^9} = \dfrac{q^{16}}{q^9} = q^{16-9} = \boxed{\bf{q^7}}$

4)

$\dfrac{\left(3^7\right)^3 \cdot 3^5}{3^{11} \cdot \left(3^6\right)^2} = \dfrac{3^{7\cdot 3}\cdot 3^5}{3^{11}\cdot 3^{6\cdot 2}} = \dfrac{3^{21}\cdot 3^5}{3^{11}\cdot 3^{12}} = \dfrac{3^{21+5}}{3^{11+12}} = \dfrac{3^{26}}{3^{23}} = 3^{26-23} = 3^3 = \boxed{\bf{27}}$