Tolong di bantu kak buat hari ini soalnya.... Question from @Jillaja

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Nomor 1.
x² - xy = 12,
x - y = 3 --> y = -3 + x
x² - x(-3 + x) = 12
x² - x² + 3x = 12
3x = 12
x = 12/3 = 4
maka y = -3 + x
y = -3 + 4 = 1

Nomor 2.
y = x² - 9 ---> y = (x + 3) (x - 3)
y = x + 3
y/y = ((x + 3) (x - 3))/ (x + 3)
1 = x - 3
x = 1 + 3 = 4
y = 4 + 3 = 7

p² + q² = 4² + 7² = 16 + 49 = 65

Nomor 3.
x² - y² = 18 ---> (x + y) (x - y) = 18
3x + 3y = 6 ---> 3 (x + y) = 6 ---> x + y = 6/3 = 2
2(x - y) = 18
x - y = 18/2 = 9

Nomor 4.
y = x² ---> √y = x
y = (2x + 3)² ---> √y = 2x + 3

x = 2x + 3
x - 2x = 3
-x = 3
x = -3

maka y = x²
y = (-3)² = 9

x + y = -3 + 9 = 6

Nomor 5.
$\frac{ {x}^{3} }{ \sqrt[3]{ x\sqrt{x} } } = {x}^{p}$
=x³ bagi (x√x)^⅓
=x³ bagi (√x.x²)^⅓
=x³ bagi (√x³)^⅓
=x³ bagi (x³^½^⅓)
=x³ bagi (x^½)
=x^(3-½)
=x^2½ ----> x^p
p= 2½

Semoga membantu and please mark as brainlist

1 votes Thanks 2

algebralover Soal No. 1
x² - xy = 12
x - y = 3

$x^{2} -xy=12 \\ x.(x-y)=12 \\ x.(3)=12 \\ \\ x= \frac{12}{3} =4 \\ \\ x-y=3 \\ 4-y=3 \\ y=4-3=1 \to Jawaban\ (C)$

Soal No. 2
y = x² - 9 dan y = x + 3
x² - 9 = x + 3
x² - x - 9 - 3 = 0
x² - x - 12 = 0
(x - 4)(x + 3) = 0
x₁ - 4 = 0
x₁ = 4
x₂ + 3 = 0
x₂ = - 3

(x₁,x₂) = (p,q)
p² + q² = 4² + (-3)² = 16 + 9 = 25...Jawaban (....?)

Soal No. 3
a² - b² = (a - b)(a + b)
x² - y² = 18
3x + 3y = 6.....dibagi dengan 3
x + y = 2

x² - y² = (x - y).(x + y)
18 = (x - y).2
(x - y) = 18 / 2
x - y = 9 ...........Jawaban (C)

Soal No. 4
y = x²

y = (2x + 3)²
x² = (2x + 3)²
x² = 4x² + (2.(2x).(3)) + 3²
x² = 4x² + 12x + 9
4x² + 12x + 9 - x² = 0
3x² + 12x + 9 = 0
(3x + 3) (x + 3) = 0
3x₁ + 3 = 0
3x₁ = - 3
x₁ = - 1

x₂ + 3 = 0
x₂ = - 3

y = (2x₁ + 3)²
y = (2.(-1) + 3)²
y = (-2 + 3)²
y = (- 1)²
y = 1

y = x₂²
y = (-3)² = 9

x₁ + y = (-1) + 1 = 0.......... Jawaban (C)
x₂ + y = (-3) + 9 = 6...........Jawaban (E)

Soal No, 5

$\frac{x^{3} }{ \sqrt[3]{x \sqrt{x} } } = x^{p} \\ \\ \frac{x^{3} }{ \sqrt[3]{(x). (x^{ \frac{1}{2} }) } } = x^{p} \\ \\ \\ \frac{x^{3} }{ (x^{ \frac{1}{3} }).( x^{ \frac{1}{6} }) } = x^{p} \\ \\ \\ \frac{x^{3} }{ x^{ \frac{1}{3}+\frac{1}{6} } } = x^{p} \\ \\ \\ \frac{x^{3} }{ x^{ \frac{1}{2} } } = x^{p} \\ \\ x^{3- \frac{1}{2} }=x^{p} \\ \\x^{2\frac{1}{2} }=x^{p} \\ \\ 2 \frac{1}{2}=p ..........Jawaban\ (D)$

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