总述:本次共解1题。其中
☆方程1题
〖 1/1方程〗
作业:求方程 7980000 = 100(20000-600m)+80(30000-400m)+(160-m)(20000+1000m) 的解.
题型:方程
解:原方程: | 7980000 | = | 100 | ( | 20000 | - | 600 | m | ) | + | 80 | ( | 30000 | - | 400 | m | ) | + | ( | 160 | - | m | ) | ( | 20000 | + | 1000 | m | ) |
去掉方程右边的括号:
方程右边 = | 100 | × | 20000 | - | 100 | × | 600 | m | + | 80 | ( | 30000 | - | 400 | m | ) | + | ( | 160 | - | m | ) | ( | 20000 | + | 1000 | m | ) |
= | 2000000 | - | 60000 | m | + | 80 | ( | 30000 | - | 400 | m | ) | + | ( | 160 | - | m | ) | ( | 20000 | + | 1000 | m | ) |
= | 2000000 | - | 60000 | m | + | 80 | × | 30000 | - | 80 | × | 400 | m | + | ( | 160 | - | m | ) | ( | 20000 | + | 1000 | m | ) |
= | 2000000 | - | 60000 | m | + | 2400000 | - | 32000 | m | + | ( | 160 | - | m | ) | ( | 20000 | + | 1000 | m | ) |
= | 4400000 | - | 92000 | m | + | ( | 160 | - | m | ) | ( | 20000 | + | 1000 | m | ) |
= | 4400000 | - | 92000 | m | + | 160 | ( | 20000 | + | 1000 | m | ) | - | m | ( | 20000 | + | 1000 | m | ) |
= | 4400000 | - | 92000 | m | + | 160 | × | 20000 | + | 160 | × | 1000 | m | - | m | ( | 20000 | + | 1000 | m | ) |
= | 4400000 | - | 92000 | m | + | 3200000 | + | 160000 | m | - | m | ( | 20000 | + | 1000 | m | ) |
= | 7600000 | + | 68000 | m | - | m | ( | 20000 | + | 1000 | m | ) |
= | 7600000 | + | 68000 | m | - | m | × | 20000 | - | m | × | 1000 | m |
= | 7600000 | + | 48000 | m | - | m | × | 1000 | m |
方程化为:
| 7980000 | = | 7600000 | + | 48000 | m | - | m | × | 1000 | m |
方程化为一般式后,用因式分解法化为:
( m - 10 )( m - 38 )=0
由
m - 10 = 0
m - 38 = 0
得:
m1=10
m2=38
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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