Soal yg amat sangad susah,kerjakan 6&7 sj.selamt mengerjakan

Berikut ini adalah pertanyaan dari ItsNotFaraz pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

Soal yg amat sangad susah,kerjakan 6&7 sj.

selamt mengerjakan

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

$\large\text{$\begin{aligned}&\textsf{6. }\ \bf\frac{1}{9}(40\sqrt{3}+39)\\\\&\textsf{7. }\ \bf{-}\frac{16}{9}\end{aligned}$}$

Pembahasan

Deret dan Eksponen

Nomor 6

$\large\text{$\begin{aligned}&\left(\sqrt{3}\right)^{-3}+\left(\sqrt{3}\right)^{-2}+\left(\sqrt{3}\right)^{-1}+\left(\sqrt{3}\right)^{0}\\&+\left(\sqrt{3}\right)^{1}+\left(\sqrt{3}\right)^{2}+\left(\sqrt{3}\right)^{3}\\{=\ }&\sum_{n={-}3}^3{\left(\sqrt{3}\right)^{n}}\\{=\ }&\bf S_7\ \begin{cases}a=U_1=\left(\sqrt{3}\right)^{-3}\\r=\sqrt{3}\end{cases}\\\end{aligned}$}$

$\large\text{$\begin{aligned}&\left[\ S_n=\frac{a(r^n-1)}{r-1}\:,\ \ r>1\right.\\\\{=\ }&\frac{\left(\sqrt{3}\right)^{-3}\left[\left(\sqrt{3}\right)^7-1\right]}{\sqrt{3}-1}\\\\{=\ }&\frac{\frac{1}{\left(\sqrt{3}\right)^{3}}\left[3^3\cdot\sqrt{3}-1\right]}{\sqrt{3}-1}\\\\{=\ }&\frac{\frac{1}{3\sqrt{3}}\left(27\sqrt{3}-1\right)}{{\sqrt{3}-1}}\\\\{=\ }&\frac{27\sqrt{3}-1}{3\sqrt{3}\left(\sqrt{3}-1\right)}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}\end{aligned}$}$

$\large\text{$\begin{aligned}{=\ }&\frac{81-\sqrt{3}+27\sqrt{3}-1}{3\sqrt{3}(3-1)}\\\\{=\ }&\frac{80+(-1+27)\sqrt{3}}{3\sqrt{3}(2)}\\\\{=\ }&\frac{80+26\sqrt{3}}{3\sqrt{3}(2)}\\\\{=\ }&\frac{40+13\sqrt{3}}{3\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}\\\\{=\ }&\bf\frac{40\sqrt{3}+39}{9}\\\\{=\ }&\bf\frac{1}{9}(40\sqrt{3}+39)\end{aligned}$}$

Nomor 7

$\large\text{$\begin{aligned}\left(\sqrt[3]{\frac{1}{243}}\right)^{3x}&=\left(\frac{3}{3^{x-2}}\right)^2\sqrt[3]{\frac{1}{9}}\\\left(243^{\frac{-1}{3}}\right)^{3x}&=\frac{3^2}{3^{2x-4}}\cdot3^{\frac{-2}{3}}\\3^{2x-4}\cdot243^{\frac{-3x}{3}}&=3^{2-\frac{2}{3}}\\3^{2x-4}\cdot\left(3^5\right)^{-x}&=3^{\frac{4}{3}}\\3^{2x-4}\cdot3^{-5x}&=3^{\frac{4}{3}}\\3^{2x-4-5x}&=3^{\frac{4}{3}}\\3^{-3x-4}&=3^{\frac{4}{3}}\\\end{aligned}$}$