Jawab:
[tex]\begin{aligned}{\sf{b.~}}&{\sf{\frac{2x+8}{x-1}}}\\{\sf{c.~}}&{\sf{\frac{5x-4}{x-2}}}\end{aligned}[/tex]
Penjelasan dengan langkah-langkah:
› Bagian b
[tex]\begin{aligned}{\sf{\frac{2x^2-32}{x^2-5x+4}}}&={\sf{\frac{2\cancel{(x-4)}(x+4)}{(x-1)\cancel{(x-4)}}}}\\&={\sf{\frac{2(x+4)}{x-1}}}\\&={\sf{\frac{2x+8}{x-1}}}\end{aligned}[/tex]
·
› Bagian c
[tex]\begin{aligned}{\sf{\frac{25x^2-16}{5x^2-6x-8}}}&={\sf{\frac{(5x-4)\cancel{(5x+4)}}{\cancel{(5x+4)}(x-2)}}}\\&={\sf{\frac{5x-4}{x-2}}}\end{aligned}[/tex]
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Verified answer
Jawab:
[tex]\begin{aligned}{\sf{b.~}}&{\sf{\frac{2x+8}{x-1}}}\\{\sf{c.~}}&{\sf{\frac{5x-4}{x-2}}}\end{aligned}[/tex]
Penjelasan dengan langkah-langkah:
› Bagian b
[tex]\begin{aligned}{\sf{\frac{2x^2-32}{x^2-5x+4}}}&={\sf{\frac{2\cancel{(x-4)}(x+4)}{(x-1)\cancel{(x-4)}}}}\\&={\sf{\frac{2(x+4)}{x-1}}}\\&={\sf{\frac{2x+8}{x-1}}}\end{aligned}[/tex]
·
› Bagian c
[tex]\begin{aligned}{\sf{\frac{25x^2-16}{5x^2-6x-8}}}&={\sf{\frac{(5x-4)\cancel{(5x+4)}}{\cancel{(5x+4)}(x-2)}}}\\&={\sf{\frac{5x-4}{x-2}}}\end{aligned}[/tex]
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