Uji Minda 1.2a – Buku Teks Matematik Tingkatan 3 Bab 1 (Indeks)


Soalan 1:
Permudahkan setiap yang berikut.
$$ \text { (a) } 3^2 \times 3 \times 3^4 $$
$$ \text { (b) }(-0.4)^4 \times(-0.4)^3 \times(-0.4) $$
$$ \text { (c) }\left(\frac{4}{7}\right) \times\left(\frac{4}{7}\right)^3 \times\left(\frac{4}{7}\right)^5 $$
$$ \text { (d) }\left(-1 \frac{2}{5}\right)^2 \times\left(-1 \frac{2}{5}\right)^3 \times\left(-1 \frac{2}{5}\right)^5 $$
$$ \text { (e) } 4 m^2 \times \frac{1}{2} m^3 \times(-3) m^4 $$
$$ \text { (f) } n^6 \times \frac{4}{25} n^2 \times \frac{5}{4} n^3 \times n $$
$$ \text { (g) }-x^4 \times \frac{25}{4} x \times \frac{12}{5} x^2 $$
$$ \text { (h) }-\frac{1}{2} y^5 \times(-6) y^3 \times \frac{1}{3} y^4 $$

Penyelesaian:
(a)
$$ \begin{aligned} & 3^2 \times 3 \times 3^4 \\ & =3^{2+1+4} \\ & =3^7 \end{aligned} $$
(b)
$$ \begin{aligned} & (-0.4)^4 \times(-0.4)^3 \times(-0.4) \\ & =(-0.4)^{4+3+1} \\ & =(-0.4)^8 \end{aligned} $$


(c)
$$ \begin{aligned} & \left(\frac{4}{7}\right) \times\left(\frac{4}{7}\right)^3 \times\left(\frac{4}{7}\right)^5 \\ & =\left(\frac{4}{7}\right)^{1+3+5} \\ & =\left(\frac{4}{7}\right)^9 \end{aligned} $$
(d)
$$ \begin{aligned} & \left(-1 \frac{2}{5}\right)^2 \times\left(-1 \frac{2}{5}\right)^3 \times\left(-1 \frac{2}{5}\right)^5 \\ & =\left(-1 \frac{2}{5}\right)^{2+3+5} \\ & =\left(-1 \frac{2}{5}\right)^{10} \end{aligned} $$


(e)
$$ \begin{aligned} & 4 m^2 \times \frac{1}{2} m^3 \times(-3) m^4 \\ = & {\left[4 \times \frac{1}{2} \times(-3)\right]\left(m^2 \times m^3 \times m^4\right) } \\ = & -6 m^{2+3+4} \\ = & -6 m^9 \end{aligned} $$


(f)
$$ \begin{aligned} & n^6 \times \frac{4}{25} n^2 \times \frac{5}{4} n^3 \times n \\ & =\left(\frac{4}{25} \times \frac{5}{4}\right)\left(n^6 \times n^2 \times n^3 \times n^1\right) \\ & =\left(\frac{1}{5}\right) n^{6+2+3+1} \\ & =\frac{n^{12}}{5} \end{aligned} $$


(g)
$$ \begin{aligned} & -x^4 \times \frac{25}{4} x \times \frac{12}{5} x^2 \\ & =\left(-1 \times \frac{25}{4} \times \frac{12}{5}\right)\left(x^4 \times x \times x^2\right) \\ & =(-15) x^{4+1+2} \\ & =-15 x^7 \end{aligned} $$


(h)
$$ \begin{aligned} & -\frac{1}{2} y^5 \times(-6) y^3 \times \frac{1}{3} y^4 \\ & =\left(-\frac{1}{2} \times-6 \times \frac{1}{3}\right)\left(y^5 \times y^3 \times y^4\right) \\ & =y^{5+3+4} \\ & =y^{12} \end{aligned} $$


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