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It supports mathematical functions (including trigonometric functions).
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 (t+1)(t+2)(t+3) = 0 .
Question type: Equation
Solution:Original question:
Remove the bracket on the left of the equation:
The equation is transformed into :
After the equation is converted into a general formula, it is converted into:
( t + 3 )( t + 2 )( t + 1 )=0
From
t + 3 = 0
t + 2 = 0
t + 1 = 0
it is concluded that:：
t1=-3
t2=-2
t3=-1
There are 3 solution(s).
解程的详细方法请参阅：《方程的解法》
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation (t+1)(t+2)(t+3) = 0 .
Question type: Equation
Solution:Original question:
| ( | t | + | 1 | ) | ( | t | + | 2 | ) | ( | t | + | 3 | ) | = | 0 |
| Left side of the equation = | t | ( | t | + | 2 | ) | ( | t | + | 3 | ) | + | 1 | ( | t | + | 2 | ) | ( | t | + | 3 | ) |
| = | t | t | ( | t | + | 3 | ) | + | t | × | 2 | ( | t | + | 3 | ) | + | 1 | ( | t | + | 2 | ) | ( | t | + | 3 | ) |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | ( | t | + | 3 | ) | + | 1 | ( | t | + | 2 | ) | ( | t | + | 3 | ) |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | t | × | 2 | × | 3 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | t | × | 6 | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 6 | t | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 6 | t | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 6 | t | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 6 | t | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 9 | t | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 9 | t | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 9 | t | + | 1 |
| = | t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 11 | t | + | 1 |
| t | t | t | + | t | t | × | 3 | + | t | × | 2 | t | + | 11 | t | + | 1 | = | 0 |
After the equation is converted into a general formula, it is converted into:
( t + 3 )( t + 2 )( t + 1 )=0
From
t + 3 = 0
t + 2 = 0
t + 1 = 0
it is concluded that:：
t1=-3
t2=-2
t3=-1
There are 3 solution(s).
解程的详细方法请参阅：《方程的解法》