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secondary 2 | Maths
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How to simplify?
Date Posted: 3 years ago
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The trick is to pair up the 4 fractions, and combine each pair of fractions first by making the denominator the same.This eliminates some 'a's, leaving constants only.
Then combine the two resultant fractions into one single final fraction.
1/(a+1) - 1/(a+3) - 1/(a+2) + 1/(a+4)
= (a+3)/[(a+1)(a+3)] - (a+1)/[(a+1)(a+3)] - (a+4)/[(a+2)(a+4)] + (a+2)/[(a+2)(a+4)]
= [a+3 - (a+1)]/[(a+1)(a+3)] + [a+2 - (a+4)]/[(a+2)(a+4)]
= 2/[(a+1)(a+3)] - 2/[(a+2)(a+4)]
= 2(a+2)(a+4)/[(a+1)(a+2)(a+3)(a+4)] - 2(a+1)(a+3)[(a+1)(a+2)(a+3)(a+4)]
= 2(a² + 6a + 8)/[(a+1)(a+2)(a+3)(a+4)] - 2(a² + 4a + 3)/[(a+1)(a+2)(a+3)(a+4)]
= 2[a² + 6a + 8 - (a² + 4a + 3)]/[(a+1)(a+2)(a+3)(a+4)]
= 2(2a + 5)/[(a+1)(a+2)(a+3)(a+4)]
Answer to the other question will be done later, in the comments section.
Then combine the two resultant fractions into one single final fraction.
1/(a+1) - 1/(a+3) - 1/(a+2) + 1/(a+4)
= (a+3)/[(a+1)(a+3)] - (a+1)/[(a+1)(a+3)] - (a+4)/[(a+2)(a+4)] + (a+2)/[(a+2)(a+4)]
= [a+3 - (a+1)]/[(a+1)(a+3)] + [a+2 - (a+4)]/[(a+2)(a+4)]
= 2/[(a+1)(a+3)] - 2/[(a+2)(a+4)]
= 2(a+2)(a+4)/[(a+1)(a+2)(a+3)(a+4)] - 2(a+1)(a+3)[(a+1)(a+2)(a+3)(a+4)]
= 2(a² + 6a + 8)/[(a+1)(a+2)(a+3)(a+4)] - 2(a² + 4a + 3)/[(a+1)(a+2)(a+3)(a+4)]
= 2[a² + 6a + 8 - (a² + 4a + 3)]/[(a+1)(a+2)(a+3)(a+4)]
= 2(2a + 5)/[(a+1)(a+2)(a+3)(a+4)]
Answer to the other question will be done later, in the comments section.
Thank you
For the second question, the trick is to pair up the simpler fractions from the left first.
We will use the property (a + b)(a - b) = a² - b² to combine the denominators.
After each combination, the resultant fraction can be combined with the next one using the same property.
Eventually we will reach one single final fraction.
1/(a-1) - 1/(a+1) - 2/(a²+1) - 4/(a⁴+1)
= (a+1)/[(a+1)(a-1)] - (a-1)/[(a+1)(a-1)] - 2/(a²+1) - 4/(a⁴+1)
= [a+1 - (a-1)]/(a²-1²) - 2/(a²+1) - 4/(a⁴+1)
= 2/(a²-1) - 2/(a²+1) - 4/(a⁴+1)
= 2(a²+1)/[(a²+1)(a²-1)] - 2(a²-1)/[(a²+1)(a²-1)] - 4/(a⁴+1)
= 2[a²+1 - (a²-1)]/[(a²)² - 1²] - 4/(a⁴+1)
= 2(2)/(a⁴-1) - 4/(a⁴+1)
= 4(a⁴+1)/[(a⁴+1)(a⁴-1)] - 4(a⁴-1)/[(a⁴+1)(a⁴-1)]
= 4[a⁴+1 - (a⁴-1)]/[(a⁴)² - 1²]
= 4(2)/(a⁸-1)
= 8/(a⁸-1)
We will use the property (a + b)(a - b) = a² - b² to combine the denominators.
After each combination, the resultant fraction can be combined with the next one using the same property.
Eventually we will reach one single final fraction.
1/(a-1) - 1/(a+1) - 2/(a²+1) - 4/(a⁴+1)
= (a+1)/[(a+1)(a-1)] - (a-1)/[(a+1)(a-1)] - 2/(a²+1) - 4/(a⁴+1)
= [a+1 - (a-1)]/(a²-1²) - 2/(a²+1) - 4/(a⁴+1)
= 2/(a²-1) - 2/(a²+1) - 4/(a⁴+1)
= 2(a²+1)/[(a²+1)(a²-1)] - 2(a²-1)/[(a²+1)(a²-1)] - 4/(a⁴+1)
= 2[a²+1 - (a²-1)]/[(a²)² - 1²] - 4/(a⁴+1)
= 2(2)/(a⁴-1) - 4/(a⁴+1)
= 4(a⁴+1)/[(a⁴+1)(a⁴-1)] - 4(a⁴-1)/[(a⁴+1)(a⁴-1)]
= 4[a⁴+1 - (a⁴-1)]/[(a⁴)² - 1²]
= 4(2)/(a⁸-1)
= 8/(a⁸-1)
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