Tolong bantuin dong :).... Question from @owk

Takamori37 Nomor 1.
49x² - 56x + 1 = 0
Sebenarnya akan sulit apabila difaktorkan.
Dengan menggunakan rumus ABC:
$\displaystyle x=\frac{-(-56)\pm\sqrt{(-56)^2-4(49)1}}{98} \\\\ x=\frac{56\pm\sqrt{3136-196}}{98} \\\\ x=\frac{56\pm\sqrt{2940}}{98}=\frac{56\pm14\sqrt{15}}{98} \\\\ x=\frac{1}{7}(4\pm\sqrt{15})$
Sehingga didapat selesaian:
HP = {x | 1/7 (4 - √15), 1/7 (4 + √15) }

Nomor 2.
$\displaystyle \frac{1}{x}+\frac{1}{x+1}=\frac{3}{2} \\\\ \frac{x+1+x}{x(x+1)}=\frac{3}{2} \\\\ \frac{2x+1}{x^2+x}=\frac{3}{2} \\\\ 2(2x+1)=3(x^2+x) \\ 4x+2=3x^2+3x \\ 3x^2-x-2=0 \\ (3x+2)(x-1)=0$
Sehingga,
HP = {x | -2/3, 1 }

1 votes Thanks 3

owk thnks

Takamori37 Iya, sama-sama.

Kto napisał słynnego krakowiaka ''Laboga Dziewczyny''???

Bantu jawab no 3 dong

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