Tolong di bantu kak buat hari ini soalnya

Tolong di bantu kak buat hari ini soalnya

Jawaban Terkonfirmasi

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

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

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.
$frac{ {x}^{3} }{ sqrt[3]{ xsqrt{x} } } = {x}^{p}$
=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

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