1.perhatikan bentuk al jabar 9y-4xy+5y+7y pangkat2+3xy-8 a.terdiri dari berapa suku bentuk al jabar tersebut?sebutkan masing2×sukunya b.sebutkan koefisien y pan

1. 9y - 4xy + 5y + 7y² + 3xy - 8
= 7y² - xy + 14y - 8
a. Terdiri dari 4 suku yaitu suku y², suku xy, suku y dan suku konstanta
b. Koefisien y² = 7
c. Koefisien xy = -4 + 3 = -1
d. Bentuk konstanta = -8

2.
a. 9p² - 4pq - q² - 4² + 5pq - 3q²
= 9p² - 4pq + 5pq - q² - 3q² - 4²
= 9p² + pq - 4q² - 16
b. 2(-8a - 3b)
= -16a - 6b
c. (2x + 3)(x - 5)
= 2x² - 10x + 3x - 15
= 2x² - 7x - 15
d. (x + 6)²
= (x + 6)(x + 6)
= x² + 6x + 6x + 36
= x² + 12x + 36

4.
[tex]a. \: \frac{y}{2} + \frac{x - 3}{3y} \\ = \frac{3y(y)}{2(3y)} + \frac{2(x - 3)}{2(3y)} \\ = \frac{3 {y}^{2} }{6y} + \frac{2x - 6}{6y} \\ = \frac{3 {y}^{2} + 2x - 6 }{6y} \\ [/tex]
[tex]b. \: \frac{x + 2}{2} - \frac{x + 3}{3} \\ = \frac{3(x + 2)}{2 \times 3} - \frac{2(x + 3)}{2 \times 3} \\ = \frac{3x + 6}{6} - \frac{2x + 3}{6} \\ = \frac{3x + 6 - (2x + 3)}{6} \\ = \frac{x + 3}{6} [/tex]
[tex]c. \: \frac{3ab}{2c} \div \frac{9 {b}^{2} }{4ac} \\ = \frac{3ab}{2c} \times \frac{4ac}{9 {b}^{2} } \\ = \frac{12 {a}^{2}bc }{18 {b}^{2} c} \\ = \frac{2 {a}^{2}bc }{3 {b}^{2} c} [/tex]

5.
a. (3x + 4) cm - (x - 2) cm
= (3x + 4) - (x - 2) cm
= 3x + 4 - x + 2
= 2x + 6 cm
b. x = 13 cm, sisa
= 2x + 6
= 2(13) + 6
= 26 + 6
= 32 cm

6. p = (5x + 2) cm, l = (3x - 4) cm
a. K = 2(p + l)
= 2((5x + 2) + (3x - 4)) cm
= 2(8x - 2) cm
= 16x - 4 cm
b. L = p × l
= (5x + 2)(3x - 4)
= 15x² - 20x + 6x - 8
= 15x² - 14x - 8
c. x = 9
K = 16x - 4
K = 16(9) - 4
K = 140 cm

L = 15x² - 14x - 8
L = 15(9²) - 14(9) - 8
L = 15(81) - 126 - 8
L = 1.215 - 134
L = 1.081 cm²