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Current location：Mathematical operation > History of Mathematical Computation > Answer
Overview: 1 questions will be solved this time.Among them
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation （10000+880a）×（1－5/18a%） = （2000+400a）×（1－3/10a%）+（8000+480a）×（1－1/4a%） .
Question type: Equation
Solution:Original question:
Remove the bracket on the left of the equation:
The equation is transformed into :
Remove the bracket on the right of the equation:
The equation is transformed into :
After the equation is converted into a general formula, it is converted into:
( a + 40 )( a - 0 )=0
From
a + 40 = 0
a - 0 = 0
it is concluded that:：
a1=-40
a2=0
There are 2 solution(s).
解一元二次方程的详细方法请参阅：《一元二次方程的解法》
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation （10000+880a）×（1－5/18a%） = （2000+400a）×（1－3/10a%）+（8000+480a）×（1－1/4a%） .
Question type: Equation
Solution:Original question:
| ( | 10000 | + | 880 | a | ) | ( | 1 | − | 5 | ÷ | 18 | × | a | ) | = | ( | 2000 | + | 400 | a | ) | ( | 1 | − | 3 | ÷ | 10 | × | a | ) | + | ( | 8000 | + | 480 | a | ) | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| Left side of the equation = | 10000 | ( | 1 | − | 5 | ÷ | 18 | × | a | ) | + | 880 | a | ( | 1 | − | 5 | ÷ | 18 | × | a | ) |
| = | 10000 | × | 1 | − | 10000 | × | 5 | ÷ | 18 | × | a | + | 880 | a | ( | 1 | − | 5 | ÷ | 18 | × | a | ) |
| = | 10000 | − | 25000 9 | a | + | 880 | a | ( | 1 | − | 5 | ÷ | 18 | × | a | ) |
| = | 10000 | − | 25000 9 | a | + | 880 | a | × | 1 | − | 880 | a | × | 5 | ÷ | 18 | × | a |
| = | 10000 | − | 25000 9 | a | + | 880 | a | − | 2200 9 | a | a |
| = | 10000 | − | 17080 9 | a | − | 2200 9 | a | a |
| 10000 | − | 17080 9 | a | − | 2200 9 | a | a | = | ( | 2000 | + | 400 | a | ) | ( | 1 | − | 3 | ÷ | 10 | × | a | ) | + | ( | 8000 | + | 480 | a | ) | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| Right side of the equation = | 2000 | ( | 1 | − | 3 | ÷ | 10 | × | a | ) | + | 400 | a | ( | 1 | − | 3 | ÷ | 10 | × | a | ) | + | ( | 8000 | + | 480 | a | ) | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 2000 | × | 1 | − | 2000 | × | 3 | ÷ | 10 | × | a | + | 400 | a | ( | 1 | − | 3 | ÷ | 10 | × | a | ) | + | ( | 8000 | + | 480 | a | ) | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 2000 | − | 600 | a | + | 400 | a | ( | 1 | − | 3 | ÷ | 10 | × | a | ) | + | ( | 8000 | + | 480 | a | ) | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 2000 | − | 600 | a | + | 400 | a | × | 1 | − | 400 | a | × | 3 | ÷ | 10 | × | a | + | ( | 8000 | + | 480 | a | ) |
| = | 2000 | − | 600 | a | + | 400 | a | − | 120 | a | a | + | ( | 8000 | + | 480 | a | ) | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 2000 | − | 200 | a | − | 120 | a | a | + | ( | 8000 | + | 480 | a | ) | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 2000 | − | 200 | a | − | 120 | a | a | + | 8000 | ( | 1 | − | 1 | ÷ | 4 | × | a | ) | + | 480 | a | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 2000 | − | 200 | a | − | 120 | a | a | + | 8000 | × | 1 | − | 8000 | × | 1 | ÷ | 4 | × | a |
| = | 2000 | − | 200 | a | − | 120 | a | a | + | 8000 | − | 2000 | a | + | 480 | a | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 10000 | − | 2200 | a | − | 120 | a | a | + | 480 | a | ( | 1 | − | 1 | ÷ | 4 | × | a | ) |
| = | 10000 | − | 2200 | a | − | 120 | a | a | + | 480 | a | × | 1 | − | 480 | a | × | 1 |
| = | 10000 | − | 2200 | a | − | 120 | a | a | + | 480 | a | − | 120 | a | a |
| = | 10000 | − | 1720 | a | − | 120 | a | a | − | 120 | a | a |
| 10000 | − | 17080 9 | a | − | 2200 9 | a | a | = | 10000 | − | 1720 | a | − | 120 | a | a | − | 120 | a | a |
After the equation is converted into a general formula, it is converted into:
( a + 40 )( a - 0 )=0
From
a + 40 = 0
a - 0 = 0
it is concluded that:：
a1=-40
a2=0
There are 2 solution(s).
解一元二次方程的详细方法请参阅：《一元二次方程的解法》