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If the equation (a + 4) x | a | - 3 + 2 = 0 is a linear equation with one variable, then the value of a is_____ .
The degree of the equation x is 1, and the coefficient of X is not equal to 0
So | a | - 3 = 1
|a|=4
a=±4
Coefficient a + 4 ≠ 0
a≠-4
So a = 4
It is known that the equation (a - 3) x | a | - 2 ← (here is x exponent) + 2 = 0 is a linear equation with one variable, and the value of a is obtained
It is known from the title that (A-3) XA = 0, A-3 = 0, a = 3
A: the value of a is 3
(I don't know if it's right, just believe me)
It is known that the | m | power + 5 = 0 of (m-1) x is a one variable linear equation about X. 1. Find the value of M. 2. Please write this equation. 3. Judge x = 1, x = 2.5, x = 3
It is known that the | m | power + 5 = 0 of (m-1) multiplied by X is a linear equation of one variable with respect to X
1. Find the value of M
Please write the equation
3. Judge whether x = 1, x = 2.5, x = 3 are solutions of the equation
It is known that the | m | power + 5 = 0 of (m-1) multiplied by X is a linear equation of one variable with respect to X
Then M-1 ≠ 0, | m | = 1
So m = - 1
The equation is: - 2x + 5 = 0
2x=5
x=2.5
So x = 2.5 is the solution of the equation, x = 1, x = 3 is not the solution of the equation
It is known that the two roots of the equation x-5x + 6 = 0 are α and β. By using the relationship between roots and coefficients, the values of α & # 179; + β & # 179; are obtained
α and β are two roots of the quadratic equation x & # 178; - 5x + 6 = 0
Then, according to the relationship between the root of quadratic equation and coefficient, α + β = 5, α β = 6
∴ α³+β³=（α+β）（α²-αβ+β²）=5*【（α+β）²-3αβ】=5*（5²-3*6）=35
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