Soal matematik SMA kk bantuin dong cara sama jawabnya doang kk 1. · x 2. : 3. 125 - 81 4. 9 + 125.... Question from @haikaldekiller

Takamori37 Nomor 1.
$$\begin{align}\sqrt[5]{81}\times\frac12\sqrt[5]{243}&=\sqrt[5]{3^4}\times\frac12\sqrt[5]{3^5} \\ &=\frac12\times3^{4/5}\times3^{5/5} \\ &=\frac12\times3\times3^{4/5} \\ &=\frac32\times3^{4/5}\end{align}$

Nomor 2.
$$\begin{align}\frac1{27}\sqrt[3]{9}:\sqrt[6]{\frac1{243}}&=\frac1{3^3}.\sqrt[3]{3^2}:\sqrt[6]{\frac{1}{3^5}} \\ &=\frac{3^{2/3}}{3^5}:3^{-5/6} \\ &=3^{\textstyle \frac23-5-(-\frac56)} \\ &=3^{\textstyle \frac46-\frac{30}{6}+\frac56} \\ &=3^{\textstyle -\frac{21}6} \\ &=3^{\textstyle -\frac72} \\ &=\frac{1}{3^{7/2}} \\ &=\frac{1}{3^3.3^{1/2}} \\ &=\frac{1}{27\sqrt3}\end{align}$

Nomor 3.
$$\begin{align}125^{2/3}-81^{3/4}&=\sqrt[3]{125^2}-\sqrt[4]{81^3} \\ &=5^2-3^3 \\ &=25-27 \\ &=-2\end{align}$

Nomor 4.
$$\begin{align}9^{1/2}+125^{2/3}&=\sqrt9+\sqrt[3]{125^2} \\ &=3+5^2 \\ &=3+25 \\ &=28\end{align}$

Nomor 5.
$$\begin{align}16^{-\frac34}+8^{-\frac23}&=\frac{1}{16^{3/4}}+\frac{1}{8^{2/3}} \\ &=\frac{1}{\sqrt[4]{16^3}}+\frac{1}{\sqrt[3]{8^2}} \\ &=\frac{1}{2^3}+\frac{1}{2^2} \\ &=\frac{1}{8}+\frac14 \\ &=\frac{1+2}8 \\ &=\frac38 \end{align}$

Nomor 6.
$$\begin{align}49^{1/2}.36^{-1/2}&=\sqrt{49}\times\frac{1}{\sqrt{36}} \\ &=7\times\frac16 \\ &=\frac76\end{align}$

Nomor 7.
$$\begin{align}\frac{25^{1/2}-8^{-2/3}}{81^{-1/3}}&=\frac{\sqrt{25}-1/\sqrt[3]{8^2}}{1/\sqrt[3]{81}} \\ &=\frac{5-1/2^2}{1/\sqrt[3]{81}} \\ &=\frac{5-1/4}{1/\sqrt[3]{81}} \\ &=\sqrt[3]{27\times3}(19/4) \\ &=\frac{19}4\times3\sqrt[3]{3} \\ &=\frac{57}{4}\sqrt[3]3\end{align}$

Nomor 8.
$$\begin{align}\frac{(3^2)^{2/3}(8^{1/2})^2}{3^2(4^3)^{1/2}9^{-1/2}}&=\frac{3^{4/3}.8}{3^2.\sqrt{4^3}.1/\sqrt9} \\ &=\frac{3^{4/3}.\!\!\not{8}}{3^2.\!\!\not{8}/3} \\ &=\frac{3^{4/3}}{3} \\ &=3^{4/3-1} \\ &=3^{1/3} \\ &=\sqrt[3]3 \end{align}$

2 votes Thanks 2

haikaldekiller yang pertama dan yang benar :D makasih kk :) 24 poin untuk anda :D

ImmanuelSinaga $\displaystyle 1)~~~\sqrt[5]{81}~\times~\frac{1}{2}\sqrt[5]{243}~=~\sqrt[5]{3^4}~\times~\frac{1}{2}\sqrt[5]{3^5}~=~3^\frac{4}{5}~~\times~~\frac{1}{2}~\times~3^\frac{5}{5} \\ ~~~~~=~\frac{3^{\frac{9}{5}}}{2}~~~atau~~~\frac{\sqrt[5]{3^9}}{2} \\ \\ 2)~~~\frac{1}{27}\sqrt[3]{9}~:~\sqrt[6]{\frac{1}{243}}~=~3^{-3}~\times~\sqrt[3]{3^2}~:~\sqrt[6]{2^{-5}} \\ \\ ~~~~~~=3^{-3}~\times~3^{\frac{2}{3}}~:~3^{-\frac{5}{6}}~=~3^{-3+\frac{2}{3}+\frac{5}{6}}~=~3^{\frac{3}{2}}~=~\sqrt[3]{2}$

$\displaystyle3)~~~125\frac{2}{3}~-~81\frac{3}{4}~~=~~(125-81)+(\frac{2}{3}-\frac{3}{4})~=~44~+~(\frac{8-9}{12}) \\\\ ~~~~~~=~~44~-~\frac{1}{12} \\ ~~~~~~~=~~43\frac{11}{12} \\ \\ 4)~~~9\frac{1}{2}~+~125\frac{2}{3}~=~(9+125)+(\frac{1}{2}~+~\frac{2}{3}) \\ \\ ~~~~~~=~~134~+~\frac{3+2}{6}~~=~~134~+~\frac{5}{6}$

$\displaystyle 5)~~~16-\frac{3}{4}+8-\frac{2}{3}~~=~~(16+8)~+~(-\frac{3}{4}-\frac{2}{3}) \\ \\ ~~~~~~~=~~24~-~\frac{9+8}{12}~=~24~-~\frac{17}{12}~=~24~-~1\frac{5}{12} \\\\ ~~~~~~~=~~(23-1)+(1-\frac{5}{12}) \\ \\ ~~~~~~~=~~22~+~\frac{7}{12} \\ \\~~~~~~~=~~22\frac{7}{12}$

$\displaystyle 7)~~~\frac{25\frac{1}{2}-8-\frac{2}{3}}{81-\frac{1}{3}}~~=~~\frac{\frac{153-48-4}{6}}{\frac{242}{3}}~~=~~\frac{101}{6}~\times~\frac{3}{242}~~=~~\frac{101}{484}$

sisanya kerjakan sendiri

0 votes Thanks 1

Tolong jawabannya kk yg baik ;) 15 poin nih 1) persamaan kuadrat baru yang akar akarnya tiga lebihnya dari akar persamaan 3x² - x - 2 = 0 adalah... 2) jika α dan β adalah akar akar persamaan kuadrat x² - x + 3 = 0, maka persamaan kuadrat yang akar akarnya adalah...

Menentukan suku ke N suku ke 32 dari deretan 3, 8, 13, 18, 23... adalah? tolong bantu kk.. besok mau testing SMA :) btw, cara nya yang lengkap ya sm penjelasaannya :D saya kurang dlm matematik., -,-

Menentukan suku ke N suku ke 32 dari deretan 3, 8, 13, 18, 23... adalah? tolong bantu kk.. besok mau testing SMA :) btw, cara nya yang lengkap ya sm penjelasaannya :D saya kurang dlm matematik., -,-

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