Maths(HKMO)
2010-01-01 9:21 pm
Suppose there are n distinct solutions of the equation
│x -│2x + 1││ = 3, find the value of n.
回答 (2)
2010-01-01 9:31 pm
✔ 最佳答案
x-|2x+1|=3 or -3|2x+1|=x-3 or x+3
2x+1=x-3,-x+3,x+3,-x-3
x=-4,2/3,2,-4/3
Actually, only x=2 and -4/3 satisfies the equation, so n=2
2010-01-04 1:58 am
x − 2x + 1 = 3
x – |2x + 1| = 3 or x – |2x + 1| = –3
x – 3 = |2x + 1| or x + 3 = |2x + 1|
x – 3 = 2x + 1 or 3 – x = 2x + 1 or x + 3 = 2x + 1 or 2x + 1 = –x – 3
x = –4 or 2/3 or –2 or –4/3
check
when x = –4 or 2/3 , x – 3 = |2x + 1| ≥ 0, no solution
when x = –2 or –4/3 , x + 3 = |2x + 1| ≥ 0, accepted
There are 2 real solutions.
The value of n is 2.
x – |2x + 1| = 3 or x – |2x + 1| = –3
x – 3 = |2x + 1| or x + 3 = |2x + 1|
x – 3 = 2x + 1 or 3 – x = 2x + 1 or x + 3 = 2x + 1 or 2x + 1 = –x – 3
x = –4 or 2/3 or –2 or –4/3
check
when x = –4 or 2/3 , x – 3 = |2x + 1| ≥ 0, no solution
when x = –2 or –4/3 , x + 3 = |2x + 1| ≥ 0, accepted
There are 2 real solutions.
The value of n is 2.
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