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    总述:本次共解1题。其中
           ☆方程1题

〖 1/1方程〗
    作业:求方程 5*(1-0.75x)*2500*(1+2.5x)+4500*7 = (4500*7+2500*5)*[1+(5/11)x] 的解.
    题型:方程
    解:原方程:
     5(1
3
4
x ) × 2500(1 +
5
2
x ) + 4500 × 7 = (4500 × 7 + 2500 × 5)(1 + (5 ÷ 11) x )
     方程左边 = 12500(1
3
4
x )(1 +
5
2
x ) + 31500
    方程化为:
     12500(1
3
4
x )(1 +
5
2
x ) + 31500 = (4500 × 7 + 2500 × 5)(1 + (5 ÷ 11) x )
    去掉方程左边的括号:
     方程左边 = 12500 × 1(1 +
5
2
x )12500 ×
3
4
x (1 +
5
2
x ) + 31500
                  = 12500(1 +
5
2
x )9375 x (1 +
5
2
x ) + 31500
                  = 12500 × 1 + 12500 ×
5
2
x 9375 x (1 +
5
2
x ) + 31500
                  = 12500 + 31250 x 9375 x (1 +
5
2
x ) + 31500
                  = 44000 + 31250 x 9375 x (1 +
5
2
x )
                  = 44000 + 31250 x 9375 x × 19375 x ×
5
2
x
                  = 44000 + 31250 x 9375 x
46875
2
x x
                  = 44000 + 21875 x
46875
2
x x
    方程化为:
     44000 + 21875 x
46875
2
x x = (4500 × 7 + 2500 × 5)(1 + (5 ÷ 11) x )
    去掉方程右边的括号:
     方程右边 = 4500 × 7(1 + (5 ÷ 11) x ) + 2500 × 5(1 + (5 ÷ 11) x )
                  = 31500(1 + (5 ÷ 11) x ) + 12500(1 + (5 ÷ 11) x )
                  = 31500 × 1 + 31500(5 ÷ 11) x + 12500(1 + (5 ÷ 11) x )
                  = 31500 + 31500(5 ÷ 11) x + 12500(1 + (5 ÷ 11) x )
                  = 31500 + 31500 × 5 ÷ 11 × x + 12500(1 + (5 ÷ 11) x )
                  = 31500 +
157500
11
x + 12500(1 + (5 ÷ 11) x )
                  = 31500 +
157500
11
x + 12500 × 1 + 12500(5 ÷ 11) x
                  = 31500 +
157500
11
x + 12500 + 12500(5 ÷ 11) x
                  = 44000 +
157500
11
x + 12500(5 ÷ 11) x
                  = 44000 +
157500
11
x + 12500 × 5 ÷ 11 × x
                  = 44000 +
157500
11
x +
62500
11
x
                  = 44000 + 20000 x
    方程化为:
     44000 + 21875 x
46875
2
x x = 44000 + 20000 x


    方程化为一般式后,用因式分解法化为:
    ( x +0 )( 25x - 2 )=0
    由
        x + 0 = 0
        25x - 2 = 0

    得:
        x1=0
        x2=
2
25
    
    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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