Write the expansion of (2 - 1/5 x)10 up to the term in x4. Hence use the expansion to find the value of P, where P = 5 − x

$$\frac{1}{2} \log_2 81 = \log_2 (81^{1/2}) = \log_2 \sqrt{81} = \log_2 9$$

Mlinganyo unakuwa:

$$\log_2 9 + \log_2 \left(x^2 - \frac{x}{3}\right) = 1$$

$$\log_2 \left(9 \left(x^2 - \frac{x}{3}\right)\right) = 1$$ $$\log_2 \left(9x^2 - 3x\right) = 1$$

$$9x^2 - 3x = 2^1$$ $$9x^2 - 3x = 2$$

$$9x^2 - 3x - 2 = 0$$

Tumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, ambapo $a=9$, $b=-3$, $c=-2$.

$$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(9)(-2)}}{2(9)}$$ $$x = \frac{3 \pm \sqrt{9 + 72}}{18}$$ $$x = \frac{3 \pm \sqrt{81}}{18}$$ $$x = \frac{3 \pm 9}{18}$$

$$x_1 = \frac{3 + 9}{18} = \frac{12}{18} = \frac{2}{3}$$ $$x_2 = \frac{3 - 9}{18} = \frac{-6}{18} = -\frac{1}{3}$$

Kwa $x = \frac{2}{3}$: $\left(\frac{2}{3}\right)^2 - \frac{1}{3}\left(\frac{2}{3}\right) = \frac{4}{9} - \frac{2}{9} = \frac{2}{9} > 0$. Hii ni halali.
Kwa $x = -\frac{1}{3}$: $\left(-\frac{1}{3}\right)^2 - \frac{1}{3}\left(-\frac{1}{3}\right) = \frac{1}{9} + \frac{1}{9} = \frac{2}{9} > 0$. Hii pia ni halali.

Thamani za $x$ ni:

$$\boxed{x = \frac{2}{3}, -\frac{1}{3}}$$

11. Andika upanuzi wa $\left(2 - \frac{1}{5}x\right)^4$ hadi neno la $x^4$.
Tutatumia Theorem ya Binomial $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
Hapa $a=2$, $b=-\frac{1}{5}x$, na $n=4$.

Neno la kwanza ($k=0$):

$$\binom{4}{0} (2)^{4-0} \left(-\frac{1}{5}x\right)^0 = 1 \cdot 2^4 \cdot 1 = 16$$

Neno la pili ($k=1$):

$$\binom{4}{1} (2)^{4-1} \left(-\frac{1}{5}x\right)^1 = 4 \cdot 2^3 \cdot \left(-\frac{1}{5}x\right) = 4 \cdot 8 \cdot \left(-\frac{1}{5}x\right) = -\frac{32}{5}x$$

Neno la tatu ($k=2$):

$$\binom{4}{2} (2)^{4-2} \left(-\frac{1}{5}x\right)^2 = 6 \cdot 2^2 \cdot \left(\frac{1}{25}x^2\right) = 6 \cdot 4 \cdot \frac{1}{25}x^2 = \frac{24}{25}x^2$$

Neno la nne ($k=3$):

$$\binom{4}{3} (2)^{4-3} \left(-\frac{1}{5}x\right)^3 = 4 \cdot 2^1 \cdot \left(-\frac{1}{125}x^3\right) = 8 \cdot \left(-\frac{1}{125}x^3\right) = -\frac{8}{125}x^3$$

Neno la tano ($k=4$):

$$\binom{4}{4} (2)^{4-4} \left(-\frac{1}{5}x\right)^4 = 1 \cdot 2^0 \cdot \left(\frac{1}{625}x^4\right) = 1 \cdot 1 \cdot \frac{1}{625}x^4 = \frac{1}{625}x^4$$

$$\boxed{\left(2 - \frac{1}{5}x\right)^4 = 16 - \frac{32}{5}x + \frac{24}{25}x^2 - \frac{8}{125}x^3 + \frac{1}{625}x^4}$$

Sehemu ya pili ya swali haionekani wazi kwenye picha.