Quiz GC!Pake cara!No Ngasal!​

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Quiz GC!Pake cara!No Ngasal!​

Quiz GC!Pake cara!No Ngasal!​

Penjelasan dengan langkah-langkah:

semoga bermanfaat jawabannya

@dinarzefanya78

Gambar Jawaban

Nomor 6

 frac{lim}{x = > infty } frac{4 {x}^{2} - 2x + 1}{3 {x}^{2} + 2 }

 frac{lim}{x = > infty } frac{4 {x}^{2} - 2 {x}^{1} + 1 {x}^{0} }{3 {x}^{2} + 0 {x}^{1} + 2 {x}^{0} }

bandingkan dengan

 frac{lim}{x = > infty } frac{a1 {x}^{m} + a2 {x}^{m - 1} + a3{x}^{m - 2} + ...... }{b1 {x}^{n} + b2 {x}^{n - 1} + b3{x}^{n - 2} + ...... }

maka didapat :

a1 = 4

a2 = -2

a3 = 1

b1 = 3

b2 = 0

b3 = 2

m = 2

n = 2

hukumnya, bila m = n atau 2 = 2 (benar) maka nilai limit tak hingganya adalah :

 = frac{a1}{b1}

 = frac{4}{3}

Nomor 7

 frac{lim}{x = > infty } (5x - 1) - sqrt{25 {x}^{ 2 } + 5x - 7}

 frac{lim}{x = > infty } sqrt{ { (5x - 1)}^{2} }- sqrt{25 {x}^{ 2 } + 5x - 7}

 frac{lim}{x = > infty } sqrt{ 25 {x}^{2} - 10x + 1}- sqrt{25 {x}^{ 2 } + 5x - 7}

bandingkan dengan

 frac{lim}{x = > infty } sqrt{ a {x}^{2} + bx + c}- sqrt{p {x}^{ 2 } + qx + r}

maka didapat :

a = 25

b = -10

c = 1

p = 25

q = 5

r = -7

hukumnya, bila a = p atau 25 = 25 (benar) maka nilai limit tak hingganya adalah :

 = frac{b - q}{2 sqrt{a} }

 = frac{ - 10- 5}{2 sqrt{25} }

 = - frac{15}{10}

 = - frac{3}{2}

Nomor 8

 frac{lim}{x = > 5} frac{ {x}^{2} - 25}{ sqrt{ {x}^{2} + 24} - 7}

 frac{lim}{x = > 5} frac{ {x}^{2} - 25}{ sqrt{ {x}^{2} + 24} - 7} times frac{sqrt{ {x}^{2} + 24} + 7}{sqrt{ {x}^{2} + 24} + 7}

 frac{lim}{x = > 5} frac{( {x}^{2} - 25)(sqrt{ {x}^{2} + 24} + 7)}{{ (sqrt{ {x}^{2} + 24})}^{2} - {7}^{2} }

 frac{lim}{x = > 5} frac{( {x}^{2} - 25)(sqrt{ {x}^{2} + 24} + 7)}{( {x}^{2} - 25) }

 frac{lim}{x = > 5} sqrt{ {x}^{2} + 24} + 7

 = sqrt{ {5}^{2} + 24} + 7

 = 7 + 7 = 14