Question

alguém me ajuda nessas duas questões por favor matemática ​

Answer 1

Resposta:

Questão 01:

a) 7$\sqrt{11}$ + 3

b) 2$\sqrt{5}$ + 4

c) $\sqrt{343} - \sqrt{28}$ = $\sqrt{49.7} - \sqrt{4.7}$ = $7\sqrt{7} - 2\sqrt{7} = 5\sqrt{7}$

d) $\sqrt{27} + \sqrt{3}$ = $3\sqrt{3} + \sqrt{3} = 4\sqrt{3}$

e) $3\sqrt{7} - \sqrt[4]{49} =$ $3 . (7)^{1/2} - (7^2)^{1/4} = 3 . (7)^{1/2} - (7)^{2/4} = 3 . (7)^{1/2} - (7)^{1/2} = 2\sqrt{7}$

f) $2\sqrt{2} + \sqrt{8} = 2\sqrt{2} + \sqrt{4.2} = 2\sqrt{2} + 2\sqrt{2} = 4\sqrt{2}$

g) $2\sqrt{7} + \sqrt{28} = 2\sqrt{7} + \sqrt{4.7} = 2\sqrt{7} + 7\sqrt{7} = 9\sqrt{7}$

h) $\sqrt{50} - \sqrt{98} = \sqrt{25.2} - \sqrt{49.2} = 5\sqrt{2} - 7\sqrt{2} = -2\sqrt{2}$

Questão 02:

a) $\sqrt{5} .\sqrt{3} =\sqrt{5.3} = \sqrt{15}$

b) $\frac{\sqrt{12} }{\sqrt{2} } = \frac{\sqrt{6}.\sqrt{2} }{\sqrt{2} } = \sqrt{6}$

c) $\sqrt[10]{5} .\sqrt[4]{2} =$ $5^{1/10}.2^{1/4}$ (não tem como juntar bases diferentes)

d) $\sqrt[3]{6} : \sqrt[4]{9} = \frac{6^{1/3}}{9^{1/4}} =\frac{(2.3)^{1/3}}{(3^2)^{1/4}} = \frac{2^{1/3}.{3^{1/3}}}{3^{1/2}} = \frac{2^{1/3}}{3^{1/6}} = 2^{1/3}.3^{-1/6}$

e) 2$\sqrt{3}$ . 3$\sqrt{2}$ . $\sqrt{5}$ = 6$\sqrt{3.2.5}$ = 6$\sqrt{30}$

f) $\sqrt[4]{a^{3}} . \sqrt[4]{a^{2}} . \sqrt{a} = \sqrt[4]{a^3.a^2}. \sqrt{a} = \sqrt[4]{a^{5}}.\sqrt{a} = a^{5/4}. a^{1/2} = a^{5.1/4.2} = a^{5/8} ou \sqrt[8]{a^5}$