mohon bantuannya : no. 10, 11, dan 12 terima kasih
Matematika
hanabistar
Pertanyaan
mohon bantuannya : no. 10, 11, dan 12
terima kasih
terima kasih
1 Jawaban
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1. Jawaban whongaliem
[tex]10) \lim_{x \to \ 2} \sqrt{ \frac{2 x^{2} + x - 6}{3 x^{2} + 4x - 4} } = \sqrt{ \frac{2 . 2^{2} + 2 - 6}{3 . 2^{2} + 4 . 2 - 4} } [/tex]
[tex]= \sqrt{ \frac{8 + 2 - 6}{12 + 8 - 4} } [/tex]
[tex]= \sqrt{ \frac{4}{16} } [/tex]
[tex]= \frac{2}{4} [/tex]
[tex]= \frac{1}{2} ....... jawaban : B[/tex]
[tex]11) \lim_{x \to \ 3} (\frac{3 x^{2} - 6}{x - 2} + \frac{2 x^{2} - 4x}{2x - 4} ) [tex]= \frac{3 . 3^{2} - 6}{3 - 2} + \frac{2 . 3^{2} - 4 . 3}{2 . 3 - 4} [/tex]
[tex]= \frac{27 - 6}{1} + \frac{18 - 12}{6 - 4} [/tex]
[tex]= 21 + \frac{6}{2} [/tex]
= 21 + 3
= 24 ..... jawaban : D
[tex]12) \lim_{x \to \ 2} \frac{ \sqrt{3 x^{2} + 8x - 3} - \sqrt{4 x^{2} + 9} }{x - 2} X \frac{ \sqrt{3 x^{2} + 8x - 3} + \sqrt{4 x^{2} -9} }{ \sqrt{3 x^{2} + 8x- 3} + \sqrt{4 x^{2} + 9} } [/tex]
[tex]= \lim_{x \to \ 2} \frac{3 x^{2} + 8x - 3 - (4 x^{2} + 9)}{(x - 2) ( \sqrt{3 x^{2} + 8x - 3} + \sqrt{4 x^{2} + 9} } [/tex]
[tex]= \frac{3 x^{2} + 8x - 3 - 4 x^{2} - 9}{(x - 2)( \sqrt{3 x^{2} + 8x - 3} + \sqrt{4 x^{2} + 9} )} [/tex]
[tex]= \frac{- x^{2} + 8x - 12}{(x - 2)( \sqrt{3 x^{2} + 8x - 3} + \sqrt{4 x^{2} + 9}) } [/tex]
[tex]= \lim_{x \to \ 2} \frac{( - x + 6)(x - 2)}{(x - 2)( \sqrt{3 x^{2} + 8x - 3} + \sqrt{4 x^{2} + 9}) } [/tex]
[tex]= \lim_{x \to \ 2} \frac{- x + 6}{ \sqrt{3 x^{2} + 8x - 3} + \sqrt{4 x^{2} + 9} } [/tex]
[tex]= \frac{- 2 + 6}{ \sqrt{3 . 2^{2} + 8 . 2 - 3} + \sqrt{4 . 2^{2} } + 9} [/tex]
[tex]= \frac{4}{ \sqrt{12 + 16 - 3} + \sqrt{16 + 9} } [/tex]
[tex]= \frac{4}{ \sqrt{25} + \sqrt{25} } [/tex]
[tex]= \frac{4}{5 + 5} [/tex]
[tex]= \frac{4}{10} [/tex]
[tex]= \frac{2}{5} ...... jawaban : D[/tex]