Simplify the following surd expression and give the answer in form of sqrt(k) where k is an integer. (3m)

Kwa neno la kwanza, $\sqrt{\frac{8}{3}}$:

$$\sqrt{\frac{8}{3}} = \frac{\sqrt{8}}{\sqrt{3}} = \frac{\sqrt{4 \times 2}}{\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{3}}$$

Ili kuondoa surd kwenye denominator, zidisha namba na denominator kwa $\sqrt{3}$:

$$\frac{2\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}$$

Kwanza, rahisiisha $\sqrt{27}$:

$$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$$

Badilisha hii kwenye neno:

$$\frac{3}{2\sqrt{27}} = \frac{3}{2(3\sqrt{3})} = \frac{3}{6\sqrt{3}} = \frac{1}{2\sqrt{3}}$$

Ili kuondoa surd kwenye denominator, zidisha namba na denominator kwa $\sqrt{3}$:

$$\frac{1}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{6}$$

Sasa tuna maneno $\frac{2\sqrt{6}}{3}$ na $\frac{\sqrt{3}}{6}$. Ili kuyajumlisha, tunahitaji denominator ya kawaida, ambayo ni 6.

$$\frac{2\sqrt{6}}{3} + \frac{\sqrt{3}}{6} = \frac{2\sqrt{6} \times 2}{3 \times 2} + \frac{\sqrt{3}}{6} = \frac{4\sqrt{6}}{6} + \frac{\sqrt{3}}{6}$$ $$= \frac{4\sqrt{6} + \sqrt{3}}{6}$$

Ikiwa usemi huu ni sawa na $\sqrt{k}$, basi kuupiga mraba kutatoa $k$:

$$k = \left(\frac{4\sqrt{6} + \sqrt{3}}{6}\right)^2$$ $$k = \frac{(4\sqrt{6})^2 + 2(4\sqrt{6})(\sqrt{3}) + (\sqrt{3})^2}{6^2}$$ $$k = \frac{(16 \times 6) + 8\sqrt{18} + 3}{36}$$ $$k = \frac{96 + 8(3\sqrt{2}) + 3}{36}$$ $$k = \frac{96 + 24\sqrt{2} + 3}{36}$$ $$k = \frac{99 + 24\sqrt{2}}{36}$$

Kwa kuwa jibu lina neno la $\sqrt{2}$, ambalo ni namba isiyo na uwiano, $k$ haiwezi kuwa namba kamili. Hii inaonyesha kuwa usemi hauwezi kuandikwa katika fomu $\sqrt{k}$ ambapo $k$ ni namba kamili.

Kwa hivyo, jibu lililorahisishwa zaidi ni:

$$\frac{4\sqrt{6} + \sqrt{3}}{6}$$