Matemáticas, pregunta formulada por anaecheverria68az, hace 11 meses

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Respuestas a la pregunta

Contestado por jorgegonzalescreyspa
0

Respuesta:

A)

8\cdot \frac{\sqrt{6}}{4}\sqrt{2}=4\sqrt{3}\quad \left(\mathrm{Decimal:\quad }\:6.92820\dots \right)

8\cdot \frac{\sqrt{6}}{4}\sqrt{2}

=\frac{\sqrt{6}\cdot \:8\sqrt{2}}{4}

=2\sqrt{6}\sqrt{2}

=2\sqrt{2\cdot \:3}\sqrt{2}

=2\sqrt{2}\sqrt{3}\sqrt{2}

=2\cdot \:2\sqrt{3}

=4\sqrt{3}

B)

4\cdot \frac{\sqrt[3]{100}}{20}\sqrt[3]{25}=\sqrt[3]{20}\quad \left(\mathrm{Decimal:\quad }\:2.71441\dots \right)\\\\4\cdot \frac{\sqrt[3]{100}}{20}\sqrt[3]{25}\\=\frac{\sqrt[3]{100}\cdot \:4\sqrt[3]{25}}{20}\\=\frac{\sqrt[3]{100}\sqrt[3]{25}}{5}\\=\frac{\sqrt[3]{4}\cdot \:5^{\frac{2}{3}+\frac{2}{3}}}{5}\\=\frac{5\sqrt[3]{4}\sqrt[3]{5}}{5}\\=\sqrt[3]{4}\sqrt[3]{5}\\=\sqrt[3]{4\cdot \:5}\\=\sqrt[3]{20}

C)

\frac{5}{2}\cdot \frac{\sqrt{10}}{\frac{1}{4}}\sqrt{5}=50\sqrt{2}\quad \left(\mathrm{Decimal:\quad }\:70.71067\dots \right)\\\frac{5}{2}\cdot \frac{\sqrt{10}}{\frac{1}{4}}\sqrt{5}\\=4\sqrt{10}\sqrt{5}\frac{5}{2}\\=\frac{5}{2}\sqrt{5\cdot \:2}\cdot \:4\sqrt{5}\\=\frac{5}{2}\sqrt{5}\sqrt{2}\cdot \:4\sqrt{5}\\\\=4\cdot \:5\sqrt{2}\frac{5}{2}\\=\frac{5\sqrt{2}\cdot \:4\cdot \:5}{2}\\=\frac{100\sqrt{2}}{2}\\=50\sqrt{2}

D)

\frac{21}{4}x\frac{\sqrt{50x^3y^5}}{\frac{7}{16}}\sqrt{2x^3y^2}=120x^4y^3\sqrt{y}\\\frac{21}{4}x\frac{\sqrt{50x^3y^5}}{\frac{7}{16}}\sqrt{2x^3y^2}\\=\frac{21}{4}\cdot \frac{80\sqrt{2}xy^2\sqrt{xy}}{7}x\sqrt{2x^3y^2}\\=\sqrt{2}\frac{21}{4}\cdot \frac{80\sqrt{2}xy^2\sqrt{xy}}{7}xxy\sqrt{x}\\=\frac{21}{4}\cdot \frac{80\sqrt{2}x\sqrt{xy}y^2}{7}\sqrt{2}x^{1+1}\sqrt{x}y\\=\frac{21}{4}\cdot \frac{80\sqrt{2}x\sqrt{xy}y^2}{7}\sqrt{2}x^2\sqrt{x}y\\

=\frac{21\cdot \:80\sqrt{2}x\sqrt{xy}y^2\sqrt{2}x\sqrt{x}y}{4\cdot \:7}

=\frac{3360x^2y^3\sqrt{x}\sqrt{xy}}{4\cdot \:7}

=\frac{3360x^2y^3\sqrt{x}\sqrt{xy}}{28}

=120x^2y^3\sqrt{x}\sqrt{xy}\\=120x^2y^3\sqrt{x}\sqrt{x}\sqrt{y}\\=120x^2xy^3\sqrt{y}\\=120y^3x^{2+1}\sqrt{y}\\=120y^3x^3\sqrt{y}

E)

2a\frac{\sqrt[5]{196m^4}}{3}a\sqrt[5]{49m^3}=\frac{2\sqrt[5]{4}\cdot \:7^{\frac{4}{5}}a^2m\sqrt[5]{m^2}}{3}\\2a\frac{\sqrt[5]{196m^4}}{3}a\sqrt[5]{49m^3}\\=2\cdot \frac{\sqrt[5]{196m^4}}{3}a^{1+1}\sqrt[5]{49m^3}\\=2\cdot \frac{\sqrt[5]{196m^4}}{3}a^2\sqrt[5]{49m^3}\\=2a^2\frac{\sqrt[5]{196}\sqrt[5]{m^4}}{3}\sqrt[5]{49m^3}\\=2\sqrt[5]{49}a^2\frac{\sqrt[5]{196}\sqrt[5]{m^4}}{3}\sqrt[5]{m^3}\\=\frac{\sqrt[5]{196}\sqrt[5]{m^4}\cdot \:2a^2\sqrt[5]{49}\sqrt[5]{m^3}}{3}\\

=\frac{2\sqrt[5]{4}\cdot \:7^{\frac{2}{5}+\frac{2}{5}}a^2\sqrt[5]{m^4}\sqrt[5]{m^3}}{3}\\=\frac{2\sqrt[5]{4}\cdot \:7^{\frac{4}{5}}a^2\sqrt[5]{m^4}\sqrt[5]{m^3}}{3}\\=\frac{2\sqrt[5]{4}\cdot \:7^{\frac{4}{5}}a^2m\sqrt[5]{m^2}}{3}

Explicación paso a paso:

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