kasih penjelasan dong biar paham​

Berikut ini adalah pertanyaan dari jdcok21 pada mata pelajaran Matematika untuk jenjang Sekolah Dasar

Kasih penjelasan dong biar paham​
kasih penjelasan dong biar paham​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Nomor 1.

Sifat-sifat operasi bilangan berpangkat :

\boxed{\boxed{\begin{array}{ll}a^b\times a^c=a^{(b+c)}&\frac{a^b}{a^c}=a^{(b-c)}\\~\\\left(a^b\right)^c=a^{(b\times c)}\end{array}}}

\frac{16^\frac{1}{2}.9^2.4^\frac{2}{3}}{16^\frac{1}{4}.9^\frac{1}{2}.4^\frac{1}{6}}

=\frac{\left(2^4\right)^\frac{1}{2}.\left(3^2\right)^2.\left(2^2\right)^\frac{2}{3}}{\left(2^4\right)^\frac{1}{4}.\left(3^2\right)^\frac{1}{2}.\left(2^2\right)^\frac{1}{6}}

=\frac{2^{\left(4\times \frac{1}{2}\right)}.3^{(2\times 2)}.2^{\left(2\times \frac{2}{3}\right)}}{2^{\left(4\times \frac{1}{4}\right)}.3^{\left(2\times \frac{1}{2}\right)}.2^{\left(2\times \frac{1}{6}\right)}}

=\frac{2^2.3^4.2^\frac{4}{3}}{2^1.3^1.2^\frac{1}{3}}

=\frac{2^{\left(2+\frac{4}{3}\right)}.3^4}{2^{\left(1+\frac{1}{3}\right)}.3^\frac{2}{3}}

=\frac{2^\frac{10}{3}.3^4}{2^\frac{4}{3}.3^1}

=2^{\left(\frac{10}{3}-\frac{4}{3}\right)}.3^{(4-1)}

=2^2.3^3

=4\times 27

=108

\therefore \boxed{\boxed{\frac{16^\frac{1}{2}.9^2.4^\frac{2}{3}}{16^\frac{1}{4}.9^\frac{1}{2}.4^\frac{1}{6}}=2^2.3^3=108}}

\\

Nomor 2.

Sifat-sifat operasi bilangan logaritma :

\boxed{\boxed{\begin{array}{l}^a\log~b+~^a\log~c=~^a\log~(b\times c)\\~\\^a\log~b-~^a\log~c=~^a\log~\frac{b}{c}\\~\\^a\log~a^b=b.^a\log~a\\~\\^a\log~a=1\end{array}}}

2a )

^2\log~48+~^5\log~50-~^2\log~3-~^5\log~2

=\left(^2\log~48-~^2\log~3\right)+\left(^5\log~50-~^5\log~2\right)

=~^2\log~\frac{48}{3}+~^5\log~\frac{50}{2}

=~^2\log~16+~^5\log~25

=~^2\log~2^4+~^5\log~5^2

=~4.^2\log~2+~2.^5\log~5

=4+2=6

\therefore \boxed{\boxed{^2\log~48+~^5\log~50-~^2\log~3-~^5\log~2=6}}

2b )

\log~45

=\log~\left(9\times 5\right)

=\log~\left(3^2\times 5\right)

=\log~3^2+\log~5

=2.\log~3+\log~5

=2.(0,477)+0,699

=0,954+0,699

=1,653

\therefore \boxed{\boxed{\log~45=1,653}}

\\

Nomor 3.

6\sqrt{27}+\frac{1}{2}\sqrt{48}-2\sqrt{243}

=6\sqrt{9\times 3}+\frac{1}{2}\sqrt{16\times 3}-2\sqrt{81\times 3}

=6\sqrt{3^2\times 3}+\frac{1}{2}\sqrt{4^2\times 3}-2\sqrt{9^2\times 3}

=\left(6\times 3\sqrt{3}\right)+\left(\frac{1}{2}\times 4\sqrt{3}\right)-\left(2\times 9\sqrt{3}\right)

=18\sqrt{3}+2\sqrt{3}-18\sqrt{3}

=(18+2-18)\sqrt{3}

=2\sqrt{3}

\therefore \boxed{\boxed{6\sqrt{27}+\frac{1}{2}\sqrt{48}-2\sqrt{243}=2\sqrt{3}}}

\\

Nomor 4.

Perkalian faktor sekawan : \boxed{(a+b)(a-b)=a^2-b^2}

\left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)

=\left(3\sqrt{2}\right)^2-\left(2\sqrt{3}\right)^2

=\left(3^2\times \left(\sqrt{2}\right)^2\right)-\left(2^2\times \left(\sqrt{3}\right)^2\right)

=(9\times 2)-(4\times 3)

=18-12

=6

\therefore \boxed{\boxed{\left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)=6}}

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Last Update: Fri, 22 Oct 21