Question

BENTUK SEDERHANA dari :
[tex]\frac{2019^{2}(2018^{2}-2017) }{(2018^{2} -1)(2018^{3} +1)} X\frac{2017^{2} (2018^{2}+2019) }{1-2018^{3} }[/tex][tex]= . . . . .[/tex]
A. -1
B. [tex]\frac{1}{2}[/tex]
C. 2017
D. 2018
E. 2019

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Answer 1

Jawaban:

A. -1

Penjelasan dengan langkah-langkah:

misalkan 2018 = c

Maka 2019 = 2018 + 1 = c + 1

2017 = 2018 - 1 = c - 1

Lanjut..

[tex] \rm\displaystyle\frac{2019^{2}(2018^{2}-2017) }{(2018^{2} -1)(2018^{3} +1)} \times \frac{2017^{2} (2018^{2}+2019) }{1-2018^{3} }[/tex]

[tex] \rm\displaystyle=\frac{(c + 1)^{2}(c^{2}-(c - 1)) }{(c^{2} -1)(c^{3} +1)} \times \frac{(c - 1)^{2} (c^{2}+(c + 1)) }{1- c^{3} }[/tex]

[tex] \rm\displaystyle=\frac{(c + 1)^{2}(c^{2}-c + 1) (c - 1)^{2} (c^{2}+c + 1) }{(c^{2} -1)(c^{3} +1)(1- c^{3}) }[/tex]

[tex]\rm\displaystyle= \frac{(c + 1)^{2}(c^{2}-c + 1) (c - 1)^{2} (c^{2}+c + 1) }{(c -1)(c + 1)(c^{3} +1)(1- c^{3}) }[/tex]

[tex] \rm\displaystyle=\frac{(c + 1)(c^{2}-c + 1) (c - 1) (c^{2}+c + 1) }{(c^{3} +1)(1- c^{3}) }[/tex]

[tex] \rm\displaystyle=\frac{(c + 1)(c^{2}-c + 1) (c - 1) (c^{2}+c + 1) }{(c + 1)(c^{2} - c +1)(1- c^{3}) }[/tex]

[tex] \rm\displaystyle=\frac{\bcancel{(c + 1)(c^{2}-c + 1)} (c - 1) (c^{2}+c + 1) }{\bcancel{(c + 1)(c^{2} - c +1)}(1- c^{3}) }[/tex]

[tex]\rm\displaystyle= \frac{ (c - 1) (c^{2}+c + 1) }{(1- c^{3}) }[/tex]

[tex] \displaystyle=\frac{ (c - 1) (c^{2}+c + 1) }{(1 - c)(1 + c +c^{2}) }[/tex]

[tex] \displaystyle=\frac{ (c - 1) (c^{2}+c + 1) }{(1 - c)(c^{2} + c + 1) }[/tex]

[tex]\rm\displaystyle=\frac{ (c - 1) \cancel{(c^{2}+c + 1)} }{(1 - c) \cancel{(c^{2} + c + 1)} }[/tex]

[tex] \rm\displaystyle=\frac{ (c - 1) }{(1- c) }[/tex]

[tex] \displaystyle=\frac{ (2017) }{(1- 2018) }[/tex]

[tex] \displaystyle=\frac{ (2017) }{( - 2017) }[/tex]

[tex] \displaystyle= \bold{ - 1}[/tex]

Jawaban A.