总述:本次共解1题。其中
☆方程1题
〖 1/1方程〗
作业:求方程 (X1+X2+X3+X4+X5+X6+X7+X8+X9+X10+X11+X12+X13+X14+X15+X16+X17+X18+X19+X20)*(X1+X2 ) = 0 的解.
题型:方程
解:原方程:| | ( | X | × | 1 | + | X | × | 2 | + | X | × | 3 | + | X | × | 4 | + | X | × | 5 | + | X | × | 6 | ) | ( | X | × | 1 | + | X | × | 2 | ) | = | 0 |
去掉方程左边的括号:
| 方程左边 = | X | × | 1 | ( | X | × | 1 | + | X | × | 2 | ) | + | X | × | 2 | ( | X | × | 1 | + | X | × | 2 | ) | + | X | × | 3 | ( | X | × | 1 | + | X | × | 2 | ) | + | X | × | 4 | ( | X | × | 1 | + | X | × | 2 | ) |
| = | X | × | 1 | X | × | 1 | + | X | × | 1 | X | × | 2 | + | X | × | 2 | ( | X | × | 1 | + | X | × | 2 | ) | + | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | ( | X | × | 1 | + | X | × | 2 | ) | + | X | × | 3 | ( | X | × | 1 | + | X | × | 2 | ) |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | × | 1 | + | X | × | 2 |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
| = | X | × | 1 | X | + | X | × | 2 | X | + | X | × | 2 | X | + | X | × | 4 | X |
方程化为一般式后,用因式分解法化为:
( X - 0 )( X - 0 )=0
由
X - 0 = 0
X - 0 = 0
得:
X1=0
X2=0
有 2个解。
解一元二次方程的详细方法请参阅:《一元二次方程的解法》
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