1/6× – 5×/10 =6× + 9​

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1/6× – 5×/10 =6× + 9​

Jawaban:

Penjelasan dengan langkah – langkah:

frac{ 1 }{ 6x } - frac{ 5x }{ 10 } =6x+9 \ 5-3xtimes 5x=6xtimes 30x+30xtimes 9 \ 5-3x^{2}times 5=6xtimes 30x+30xtimes 9 \ 5-15x^{2}=6xtimes 30x+30xtimes 9 \ 5-15x^{2}=6x^{2}times 30+3 \ 0xtimes 9 \ 5-15x^{2}=180x^{2}+30xtimes 9 \ 5-15x^{2}=180x^{2}+270x \ 5-15x^{2}-180x^{2}=270x \ 5-15x^{2}-180x^{2}-270x=0 \ -15x^{2}-180x^{2}-270x=-5 \ -195x^{2}-270x=-5 \ frac{-195x^{2}-270x}{-195}=frac{-5}{-195} \ x^{2}+frac{-270}{-195}x=frac{-5}{-195} \ x^{2}+frac{18}{13}x=frac{-5}{-195} \ x^{2}+frac{18}{13}x=frac{1}{39} \ x^{2}+frac{18}{13}x+left(frac{9}{13}right)^{2}=frac{1}{39}+left(frac{9}{13}right)^{2} \ x^{2}+frac{18}{13}x+frac{81}{169}=frac{1}{39}+frac{81}{169} \ x^{2}+frac{18}{13}x+frac{81}{169}=frac{256}{507} \ left(x+frac{9}{13}right)^{2}=frac{256}{507} \ sqrt{left(x+frac{9}{13}right)^{2}}=sqrt{frac{256}{507}} \ x+frac{9}{13}=frac{16sqrt{3}}{39} \ x+frac{9}{13}=-frac{16sqrt{3}}{39} \ x=frac{16sqrt{3}}{39}-frac{9}{13} \ x=-frac{16sqrt{3}}{39}-frac{9}{13} \

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Gambar Jawaban

Penjelasan dengan langkah-langkah:

Pertama, ubah atau sederhanakan bentuk persamaan :

frac{1}{6x} - frac{5x}{10} = 6x + 9

frac{1}{6x} - frac{x}{2} = 6x + 9

frac{(1 times 2) - (6x times x)}{6x times 2} = 6x + 9

frac{2 - 6 {x}^{2} }{12x} = 6x + 9

2 - 6 {x}^{2} = 12x(6x + 9)

2 - 6 {x}^{2} = 72 {x}^{2} + 108x

(72 {x}^{2} + 6 {x}^{2} ) + 108x - 2 = 0

78 {x}^{2} + 108x - 2 = 0

39 {x}^{2} + 54x - 1 = 0 : (setelah : dibagi : 2)

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Tentukan nilai abc dari persamaan kuadrat tersebut :

a = 39

b = 54

c = -1

Kedua, hitung nilai akar-akar dari persamaan kuadrat tersebut :

x1,x2 = frac{ - b± sqrt{ {b}^{2} - 4ac} }{2a}

x1,x2 = frac{ - 54± sqrt{ {54}^{2} - 4.39. - 1} }{2.39}

x1,x2 = frac{ - 54± sqrt{2916 + 156} }{78}

x1,x2 = frac{ - 54± sqrt{3072} }{78}

x1,x2 = frac{ - 54± 32sqrt{3} }{78}

x1,x2 = frac{ - 27± 16sqrt{3} }{39}

x1= frac{ - 27 + 16sqrt{3} }{39}

x2 = frac{ - 27 - 16sqrt{3} }{39}