Latihan Soal Saintek

Pembahasan :

L = 96 cm²

2x² + 2xt + 2xt = 96

32 = 2x²

32 = 2xt

32 = 2xt

2x² = 32 maka x = 4

2 . 4 . t =32 maka t = 4

V = 4 . 4 . 4

= 64 cm³

$$\left|\ 8\ -\ 2x^2\ \right|>\sqrt[]{8\ -\ 2x^2}+\ 2$$

Syarat : $$8\ -\ 2x^2\ \ge\ 0$$

$$-2x^2\ \le\ -8$$

$$x\ \le\ 4$$

$$-2\ <\ x\ <\ 2$$

$$\left|\ 8\ -\ 2x^2\right|\ =\ \sqrt[]{8\ -\ 2x^2\ +\ 2}$$

$$\sqrt[]{\left(8\ -\ 2x^2\right)}^2\ >\sqrt[]{8\ -\ 2x^2}+\ 2\$$

$$p\ =\ \sqrt[]{8\ -\ 2x^2}$$

$$p^2\ >\ p\ +\ 2$$

$$\left(p\ +\ 1\right)\left(p\ -\ 2\right)>\ 0$$

$$p\ <\ -1$$

$$\sqrt[]{8\ -\ 2x^2}\ <\ -1$$

$$p\ >\ 2$$

$$\sqrt[]{8\ -\ 2x^2}>\ 2$$

$$8\ -\ 2x^2\ >\ 4$$

$$-2x^2\ >\ -4\$$

$$x^2\ <\ 2$$

$$-\sqrt[]{2}\ <\ x\ <\ \sqrt[]{2}$$

x⁴ - 2x² + 1 = ( x² - 1 )( x² - 1 )

$$\left(x^2\ -\ 1\right)\ \left(x^2\ -\ 1\right)\ =\ \left(\ x^2\ +\ ax\ +\ b\right)\ \left(x^2\ +\ cx\ +\ d\right)$$

maka :

$$x^2\ +\ ax\ +\ b\ =\ x^2\ -\ 1$$

$$x^2\ +\ cx\ +\ d\ =\ x^2\ -\ 1$$

maka ; a = 0 b = -1 c = 0 d = -1

a + b + c + d = -2

$$p\ \left(x\right)\ =\ k\left(x-1\right)\left(x\ -2\right)\left(x\ -3\right)+x\ +\ 1$$

$$p\ \left(4\right)\ =k\left(4\ -1\right)\left(4\ -2\right)\left(4\ -3\right)\ +\ 4\ +\ 1$$

$$=k\ \left(6\right)\$$

$$6k\ =\ 1$$

$$k=\ \frac{1}{6}$$

$$p\ \left(x\ +\ 2\right)\ =\ \frac{1}{6}\ \left(\ x+1\right)\ \left(\times\right)\ \left(\ x\ +\ 1\right)\ +\ x\ +\ 3$$

$$=\frac{\left(\text{(x + 1) (x) (x}+1\right)\ +\ 6x\ +\ 18}{\text{6}}$$

$$=\frac{x^3\ -\ x\ +\ 6x\ +\ 18}{6}$$

$$=\ \frac{x^3\ -5x\ +\ 18}{6}$$

$$=\frac{\left(x\ +\ 2\right)\left(x^2\ -\ 2x\ -\ 9\right)}{6}$$

$$=x\ +\ 2$$

$$AB\ =\ \sqrt[]{6^2\ +\ 8^2}$$

$$=\ 10$$

Pembahasan :

$$\overline{X_A}\ :\ \overline{X_B}\ =\ 10\ :\ 9\ =\ 90\ :\ 81$$

$$\overline{X_A}\ :\ \overline{X_B}\ =\ 85\ :\ 81\ =\ 85\ :\ 81$$

$$\overline{X_A}\ :\ \overline{X_B}\ :\ \overline{x}\ =\ 90\ :\ 81\ :\ 85$$

$$n_A\ :\ n_B\ =\ \left|85\ -\ 81\right|\ :\ \left|85\ -90\right|\$$

$$=\ 4\ :\ 5$$

Pembahasan :

r = 2 $$a\ =3\left(2^{1+1}-2\right)$$

$$=\ 6$$

Maka, 2 + 6 = 8

$$r\ =\ \sqrt[]{3^2\ +\ 4^2}\ =\ 5$$

$$M_{ab}\ =\ \frac{\text{6 - 5}}{\text{-8 - 2}}\ =\ -\frac{\text{1}}{\text{10}}$$

$$-\frac{\text{a}}{\text{b}}\ \left(-\frac{\text{1}}{\text{10}}\right)\ =\ -1$$

$$-\frac{\text{a}}{\text{b}}\ =\ 10$$

$$\left(f\ o\ g\right)^{-1}\ \left(p\ \right)=\frac{\text{3}}{\text{2}}$$

$$\left(f\ o\ g\right)^{-1}\ \left(\frac{\text{3}}{\text{2}}\right)\ =\ p$$

$$g\left(\frac{\text{3}}{\text{2}}\right)\ =\ 2\ \left(\frac{\text{3}}{\text{2}}\right)\ +\ 1$$

$$g\left(\frac{\text{3}}{\text{2}}\right)\ =4$$

$$f\ \left(g\frac{\text{3}}{\text{2}}\right)\ =\ f\ \left(4\right)\$$

$$=\frac{\text{2 (4) + 1}}{\text{4 - 3}}$$

$$=9$$