Gambar 1 nomor 1,2 dan 3 gambar 2 nomor 4,5,6 dan 7 => nomor 7 ada a dan b nya di gambar 3 gambar 3 .... Question from @steven123456

September 2018 1 96 Report

Gambar 1 nomor 1,2 dan 3
gambar 2 nomor 4,5,6 dan 7 => nomor 7 ada a dan b nya di gambar 3
gambar 3 nomor 8,9 dan 10
gambar 4 nomor 11 dan 12
klau gk jelas silahkan tnya di comentar

Takamori37 Nomor 1.
a.
$\sqrt{48}=\sqrt{16\times 3}=4\sqrt{3}$
b.
$\sqrt{54}=\sqrt{9\times 6}=3\sqrt{6}$
c.
$\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$
d.
$\sqrt{80}=\sqrt{16\times 5}=4\sqrt{5}$

Nomor 2.
Berada pada lampiran (Berhubung angkanya tidak sederhana)

Nomor 3.
a.
$$\begin{align}\frac{3}{\sqrt{6}}&=\frac{3}{\sqrt{6}}\times\frac{\sqrt{6}}{\sqrt{6}} \\ &=\frac{3\sqrt{6}}{6} \\ &=\frac{1}{2}\sqrt{6}\end{align}$
b.
$$\begin{align}\frac{3}{2\sqrt{5}}&=\frac{3}{2\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}} \\ &=\frac{3\sqrt{5}}{2\times5} \\ &=\frac{3}{10}\sqrt{5}\end{align}$
c.
$$\begin{align}\frac{8}{3+\sqrt{21}}&=\frac{8}{3+\sqrt{21}}\times\frac{\sqrt{21}-3}{\sqrt{21}-3} \\ &=\frac{8(\sqrt{21}-3)}{21-9} \\ &=\frac{8}{12}(\sqrt{21}-3) \\ &=\frac{2}{3}(\sqrt{21}-3) \end{align}$
d.
$$\begin{align}\frac{4}{7-\sqrt{11}}&=\frac{4}{7-\sqrt{11}}\times\frac{7+\sqrt{11}}{7+\sqrt{11}} \\ &=\frac{4(7+\sqrt{11})}{49-11} \\ &=\frac{4}{35}(7+\sqrt{11})\end{align}$

Nomor 4.
a.
$$\begin{align}32^{1/10}&=(2^5)^{1/10} \\ &=2^{5/10} \\ &=2^{1/2} \\ &=\sqrt{2}\end{align}$
b.
$$\begin{align}27^{5/6}&=(3^3)^{5/6} \\ &=3^{5/2} \\ &=3^2\times3^{1/2} \\ &=9\sqrt{3}\end{align}$

Nomor 5.
a.
$$\begin{align}7^{\frac{2}{3}}\times7^{\frac{1}{2}}&=7^{\frac{2}{3}+\frac{1}{2}} \\ &=7^{\frac{7}{6}}\end{align}$
b.
$$\begin{align}7^{-\frac{1}{2}}\times7^{\frac{3}{2}}&=7^{-\frac{1}{2}+\frac{3}{2}} \\ &=7^{\frac{2}{2}} \\ &=7^1 \\ &=7\end{align}$
c.
$$\begin{align}(-5)\times(-5)^{\frac{2}{3}}&=(-5)^{1+\frac{2}{3}} \\ &=(-5)^{\frac{5}{3}}\end{align}$
d.
$$\begin{align}8^{-\frac{1}{4}}\times2^{-\frac{3}{2}}&=2^{-\frac{3}{4}-\frac{3}{2}} \\ &=2^{-\frac{9}{4}}\end{align}$

Nomor 6.
$$\begin{align}\left(\frac{1}{a^3}-\frac{1}{b^3}\right)^{-\frac{1}{3}}&=\left(\frac{b^3-a^3}{a^3b^3}\right)^{-\frac{1}{3}} \\ &=\left(\frac{b^3-a^3}{a^3b^3}\right)^{\frac{1}{3}} \\ &=\frac{\sqrt[3]{b^3-a^3}}{ab}\end{align}$

Nomor 7.
Rasionalkan, didapat:
$\displaystyle p=-\frac{3}{5}(\sqrt{2}+3)$
$\displaystyle q=-\frac{3}{5}(\sqrt{2}-3)$
Maka,
$p+q=-\frac{3}{5}(\sqrt{2}+3)+(-\frac{3}{5}(\sqrt{2}-3)) \\ p+q=-\frac{6}{5}\sqrt{2}$

$p+q=-\frac{3}{5}(\sqrt{2}+3)-(-\frac{3}{5}(\sqrt{2}-3)) \\ p+q=-\frac{18}{5}\sqrt{2}$

Untuk perkalian berbeda, dari angka awalnya saja.
$p\times q=-\frac{6}{5}$

b.
Sama dengan cara bagian a.
Didapat:
p + q = 3/2 √33
p - q = 9√3
Mungkin dapat diperbaiki jika terjadi kesalahan umum.

Nomor 8.
$$\begin{align}\frac{x+\sqrt{4x^2}}{3+\sqrt{2}}&=\frac{1}{4+2\sqrt{2}} \\ 3x&=\frac{3+\sqrt{2}}{4+2\sqrt{2}} \\ x&=\frac{1}{6}\times\frac{3+\sqrt{2}}{2+\sqrt{2}} \\ x&=\frac{1}{6}\times\frac{3+\sqrt{2}}{2+\sqrt{2}}\times\frac{2-\sqrt{2}}{2-\sqrt{2}} \\ x&=\frac{1}{6}\times\frac{4+\sqrt{2}}{2} \\ x&=\frac{4+\sqrt{2}}{12} \end{align}$

Nomor 9.
$$\begin{align}\left(x^2+\frac{1}{x^2}\right)&=\left(x+\frac{1}{x}\right)^2-2 \\ &=\left(\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}\right)^2-2 \\ &=\left(\frac{6+2\sqrt{5}+6-2\sqrt{5}}{4}\right)^2-2 \\ &=3^2-2 \\ &=9-2 \\ &=7\end{align}$

Nomor 10.
Merupakan Mathematical Fallacy dengan a² di dalam akar seharusnya tidak diperbolehkan menjadi (-a)(-a).

Nomor 11.
Beri daerah tinggi atas dan tinggi bawah
Tinggi atas,
Didapat phytagoras,
√5

Tinggi bawah,
Didapat phytagoras,
4√2

Yang menyebabkan tinggi total : (4√2 + √5) cm

Nomor 12.
Dengan merasionalkan,
$\displaystyle a=3+2\sqrt{2} \\ b=3-2\sqrt{2}$
Sehingga,
$$\begin{align}a^2-b^2&=(a+b)(a-b) \\ &=(3+2\sqrt{2}+3-\sqrt{2})(3+2\sqrt{2}-(3-2\sqrt{2}) \\ &=6\times 4\sqrt{2} \\ &=24\sqrt{2}\end{align}$

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