Overview: 6 questions will be solved this time.Among them
☆3 inequalities
☆2 equations
☆1 arithmetic calculations
[1/6 Formula]
Work: Calculate the value of equation (-2)^2+6/(-3)-sqrt(9)+(-0.1)^0 .
Question type: Mathematical calculation
Solution:
(-2)^2+6/(-3)-sqrt(9)+(-0.1)^0
=-2^2+6/(-3)-sqrt(9)+(-0.1)^0
=-2^2+6/(-3)-sqrt(9)+(-0.1)^0
=-2^2+6/(-3)-sqrt9+(-0.1)^0
=-2^2+6/(-3)-sqrt9+(-0.1)^0
=-2^2+6/-3-3+-0.1^0
=4+6/(-3)-3+(-0.1)^0
=4+6/(-3)-3+1
=4+(-2)-3+1
=2-3+1
=-1+1
=0 Answer:(-2)^2+6/(-3)-sqrt(9)+(-0.1)^0=0[ 2/6Inequality]
Assignment:Find the solution set of inequality x^2-x ≤2 .
Question type: Inequality
Solution:
The inequality can be reduced to 1 inequality:
x ^ 2 - x ≤2 (1)
From inequality(1):
-1 ≤ x ≤ 2
The final solution set is :
-1 ≤ x ≤ 2[ 3/6 Equation]
Work: Find the solution of equation x^3-2*x = 1 .
Question type: Equation
Solution:
After the equation is converted into a general formula, there is a common factor:
( x + 1 )
From
x + 1 = 0
it is concluded that::
x1=-1
Solutions that cannot be obtained by factorization:
x2≈-0.618034 , keep 6 decimal places
x3≈1.618034 , keep 6 decimal places
There are 3 solution(s).
解程的详细方法请参阅:《方程的解法》[ 4/6 Equation]
Work: Find the solution of equation x^ln(x) = 4 .
Question type: Equation
Solution:
The solution of the equation:
x1≈0.308076 , keep 6 decimal places
x2≈3.245956 , keep 6 decimal places
There are 2 solution(s).
解程的详细方法请参阅:《方程的解法》[ 5/6Inequality]
Assignment:Find the solution set of inequality x+2 ≥2*x .
Question type: Inequality
Solution:
The inequality can be reduced to 1 inequality:
x + 2 ≥2 * x (1)
From inequality(1):
x ≤ 2
The final solution set is :
x ≤ 2[ 6/6Inequality]
Assignment:Find the solution set of inequality 1/x ≤1 .
Question type: Inequality
Solution:
The inequality can be reduced to 1 inequality:
1 / x ≤1 (1)
From the definition field of divisor
x ≠ 0 (2 )
From inequality(1):
x ≤ 0 或 x ≥ 1
From inequality(2):
x < 0 或 x > 0
From inequalities (1) and (2)
x < 0 或 x ≥ 1 (3)
The final solution set is :
x < 0 或 x ≥ 1Your problem has not been solved here? Please go to the Hot Problems section!