Matemática, perguntado por joao226oliveira, 7 meses atrás

Calcule as seguintes potenciações

a) (3 xym^8)²

b)(-5/6yz)³

c)[(-2x)^4]²

d)[(-2/5x^7)²]²


joao226oliveira: ME AJUDAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
joao226oliveira: favor ;-;

Soluções para a tarefa

Respondido por wlima130501
4

Resposta:

a)9x^{2}y^{2}m^{16}

b)-\frac{125}{216}y^{3}z ^{3}

c)256x^{8}

d)\frac{16}{625}x^{28}

Explicação passo-a-passo:

a)(3xym^{8})^{2} = 3^{2}x^{2}y^{2}(m^{8})^{2}=9x^{2}y^{2}m^{16}

b)(-\frac{5}{6}yz) ^{3} =(-\frac{5}{6})^{3}y^{3}z ^{3} =-\frac{125}{216}y^{3}z ^{3}

c)[(-2x)^{4}]^{2}=[(-2)^{4}x^{4}]^{2}=[16x^{4}]^{2}=16^{2}(x^{4})^{2}=256x^{8}

d)[(-\frac{2}{5}x^{7})^{2}]^{2}=[(-\frac{2}{5})^{2}(x^{7})^{2}]^{2}=[\frac{4}{25}x^{14}]^{2}=(\frac{4}{25})^{2}(x^{14})^{2}=\frac{16}{625}x^{28}


joao226oliveira: obrigado meu queridoooo
joao226oliveira: \o/
wlima130501: Por nada bem
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