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    Current location:Equations > Monovariate Equation > The history of univariate equation calculation > Answer

    Overview: 1 questions will be solved this time.Among them
           ☆1 equations

[ 1/1 Equation]
    Work: Find the solution of equation (a+1)(a÷5+1)(a÷25+1)+15 = 0 .
    Question type: Equation
    Solution:Original question:
     ( a + 1)( a ÷ 5 + 1)( a ÷ 25 + 1) + 15 = 0
    Remove the bracket on the left of the equation:
     Left side of the equation = a ( a ÷ 5 + 1)( a ÷ 25 + 1) + 1( a ÷ 5 + 1)( a ÷ 25 + 1) + 15
                                             = a a ÷ 5 × ( a ÷ 25 + 1) + a × 1( a ÷ 25 + 1) + 1( a ÷ 5 + 1)( a ÷ 25 + 1) + 15
                                             = a a ×
1
5
a ÷ 25 + a a ×
1
5
× 1 + a × 1( a ÷ 25 + 1)
                                             = a a ×
1
125
a + a a ×
1
5
+ a × 1( a ÷ 25 + 1) + 1( a ÷ 5 + 1)
                                             = a a ×
1
125
a + a a ×
1
5
+ a × 1 a ÷ 25 + a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a + a × 1
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a + 1 a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a + 1 a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a + 1 a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a + 1 a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a +
6
5
a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a +
6
5
a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a +
6
5
a
                                             = a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a +
31
25
a
    The equation is transformed into :
      a a ×
1
125
a + a a ×
1
5
+ a ×
1
25
a +
31
25
a = 0

    
        a≈-28.015548 , keep 6 decimal places
    
    There are 1 solution(s).


解一元一次方程的详细方法请参阅:《一元一次方程的解法》



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