1 Fill in the following blanks with correct answers
(1) \(2\sqrt{12} \:-\:3\sqrt{6} \div \sqrt{18}\)
\(2\sqrt{4 \times 3} \:-\:3\sqrt{6} \div \sqrt{9 \times 2}\)
\(4\sqrt{3} \:-\:3\sqrt{6} \div 3\sqrt{2}\)
\(4\sqrt{3} \:-\:\sqrt{6} \div \sqrt{2}\)
\(4\sqrt{3} \:-\:\sqrt{3}\)
\(3\sqrt{3}\)
(2) \(\dfrac{x^2 \:-\:x \:-\:6}{x^2 + x \:-\:2} \:-\: \dfrac{2x\:-\:4}{x\:-\:1}\)
\(\dfrac{x^2 \:-\:x \:-\:6}{(x + 2)(x\:-\:1)} \:-\: \dfrac{2x\:-\:4}{x\:-\:1}\)
\(\dfrac{(x^2 \:-\:x \:-\:6)\:-\:(x + 2)(2x \:-\:4)}{(x + 2)(x\:-\:1)}\)
\(\dfrac{x^2 \:-\:x \:-\:6 \:-\:(2x^2 \:-\:\cancel{4x} + \cancel{4x} \:-\:8)}{(x + 2)(x\:-\:1)}\)
\(\dfrac{-x^2\:-\:x + 2}{x^2 + x\:-\:2}\)
\(\dfrac{-\cancel{(x^2 + x \:-\: 2)}}{\cancel{x^2 + x\:-\:2}}\)
\(-1\)
2 On the plane \(xy\), there are three points: O (0, 0), A (2, 4), B (3, 0). Fill in the following blanks with correct answers.
(1) Taking point D in the fourth quadrant and when the quadrilateral ODBA is parallelogram, the coordinates of point D is…
\(\textbf{D}(1, -4)\)
(2) When a straight line \(x = p\) bisects the area of \(\triangle \text{OAB}\), \(p = \dotso\)
Area of triangle OAB = \(\dfrac{1}{2} \cdot 3 \cdot 4\)
Area of triangle OAB = 6
Persamaan garis yang melalui titik (0, 0) dan (2, 4) adalah
\(\dfrac{y\:-\:y_1}{y_2 \:-\:y_1} = \dfrac{x \:-\:x_1}{x_2 \:-\:x_1}\)
\(\dfrac{y\:-\:0}{4 \:-\:0} = \dfrac{x \:-\:0}{2 \:-\:0}\)
\(\dfrac{y}{4} = \dfrac{x}{2}\)
\(y = 2x\)
Untuk \(x = p\) maka \(y = 2p\)
Garis \(x = p\) membagi luas segitiga OAB menjadi dua sama besar.
Luas setengah segitiga OAB = 3
\(\dfrac{1}{2} \cdot p \cdot 2p = 3\)
\(p^2 = 3\)
\(p = \sqrt{3}\)
(3) When a straight line \(y = q\) bisects the area of \(\triangle \text{OAB}\), \(q = \dotso\)
(4) When the straight line L drawn from point B bisects the area of \(\triangle \text{OAB}\), the equation of the straight line L is \(y = \dotso\)
(5) The equation of a parabola which passes three points: O, A, B is \(y = \dotso\)
