Overview: 5 questions will be solved this time.Among them
☆5 equations
[ 1/5 Equation]
Work: Find the solution of equation 5(x+3) = -3x .
Question type: Equation
Solution:Original question: Remove the bracket on the left of the equation:
| Left side of the equation = | 5 | x | + | 5 | × | 3 |
The equation is transformed into :
Transposition :
Combine the items on the left of the equation:
The coefficient of the unknown number is reduced to 1 :
We obtained :
This is the solution of the equation.
Convert the result to decimal form :
[ 2/5 Equation]
Work: Find the solution of equation -(x/3) = (2x/5)+(√3)π .
Question type: Equation
Solution:
x=0
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》[ 3/5 Equation]
Work: Find the solution of equation 5x = 5/x .
Question type: Equation
Solution:Original question:| Multiply both sides of the equation by: | x |
After the equation is converted into a general formula, it is converted into:
( x + 1 )( x - 1 )=0
From
x + 1 = 0
x - 1 = 0
it is concluded that::
x1=-1
x2=1
There are 2 solution(s).
解一元二次方程的详细方法请参阅:《一元二次方程的解法》[ 4/5 Equation]
Work: Find the solution of equation x^2-4x = -3 .
Question type: Equation
Solution:
After the equation is converted into a general formula, it is converted into:
( x - 1 )( x - 3 )=0
From
x - 1 = 0
x - 3 = 0
it is concluded that::
x1=1
x2=3
There are 2 solution(s).
解一元二次方程的详细方法请参阅:《一元二次方程的解法》[ 5/5 Equation]
Work: Find the solution of equation x^4+(-1) = 0 .
Question type: Equation
Solution:
After the equation is converted into a general formula, it is converted into:
( x + 1 )( x - 1 )=0
From
x + 1 = 0
x - 1 = 0
it is concluded that::
x1=-1
x2=1
There are 2 solution(s).
解一元二次方程的详细方法请参阅:《一元二次方程的解法》Your problem has not been solved here? Please go to the Hot Problems section!