Question

Tolong dijawab ya. asal jgn salah. pake caranya. besok dikumpul. jdi buruan ya

Answer 1

(5x-2)^2 = (5x-2) (5x-2) = 5x(5x-2) -2(5x-2) =25x^2-10x-10x+4 =25x^2-20x+4 (untuk soal nomor 2)

Answer 2

1.)
(3a-5)(4a+7)
= (3a)(4a) + (3a)(7) + (-5)(4a) + (-5)(7)
= 12a² + 21a - 20a - 35
= 12a² + a - 35 [D]

2.)
(5x-2)²
= (5x-2)(5x-2)
= (5x)(5x) + (5x)(-2) + (-2)(5x) + (-2)(-2)
= 25x² - 10x - 10x + 4
= 25x² - 20x + 4 [B]

3.)
(2a+1)(a-8) + 6a² - 4
= 2a² - 16a + a - 8 + 6a² - 4
= (2a² + 6a²) + (a - 16a) - 8 - 4
= 8a² -15a - 12 [D]

4.)
3x² - 25x + 28
p + q = -25
pq = 84
Didapat, p = -21, q = -4
Menyebabkan:
= 1/3 (3x-21)(3x-4)
= (x-7)(3x-4) [B]

5 dan 6)
Melalui sifat a²-b² = (a-b)(a+b)
x² - 25 = (x-5)(x+5) [D]
9p²-16 = (3p-4)(3p+4) [C]

7.)
(4y-11)²
= (4y)² - 2(4y)(11) + (11)²
= 16y² - 88y + 121 [B]

8.)
Karena demikian:
Tanda koefisien ab adalah negatif, tengahnya juga negatif.
Koefisien a² diakarkuadratkan menjadi 2/7
Koefisen b² diakarkuadratkan menjadi 3/5
Maka, jabarannya adalah:
(2/7 a - 3/5 b)² [A]

9.)
Faktorkan menjadi:
= 6(2x+1) / [(2x+1)(2x-1)]
= 6 / [2x - 1] [C]

10.)
[2a + 10] / [3a² + 14a - 5]
= 2(a+5) / [(a+5)(3a - 1)]
= 2 / [3a - 1] [B]

11.)
1/[z+5] - [z+2] / [2z²+7z-15]
= 1/[z+5] - [z+2]/[(2z-3)(z+5)]
= 1/[z+5] x [1 - [z+2]/[2z-3]]
= 1/[z+5] x [(2z-3)-(z+2)]/[2z-3]
= 1/[z+5] x [z+5]/[2z-3]
= 1/[2z-3]