Given 2x-3 = 0, find the value of X (X & # 179; - x) + X & # 178; (5-x) - 9

Given 2x-3 = 0, find the value of X (X & # 179; - x) + X & # 178; (5-x) - 9

x﹙x³－x﹚＋x²﹙5－x﹚－9
＝x^4－x²＋5x²－x³－9
＝x^4－x³＋4x²－9
＝x^4－x³＋﹙2x－3﹚﹙2x＋3﹚
＝x³﹙x－1﹚
＝﹙3/2﹚²×﹙3/2－1﹚
＝27/16
-1 / 16 question: is it OK to write down the process? Thank you
Given the set a = {- 4, 2a-1, A2}, B = {a-5, 1-A, 9}, find the value of a which is suitable for the following conditions: (1) 9 ∈ (a ∩ b); (2) {9} = a ∩ B
(1) ∩ 9 ∈ (a ∩ b), ∩ 9 ∈ B and 9 ∈ a, ∩ 2a-1 = 9 or A2 = 9, ∩ a = 5 or a = ± 3. The test shows that a = 5 or a = - 3. (2) ∩ 9} = a ∩ B, ∩ 9 ∈ (a ∩ b), ∩ a = 5 or a = - 3. When a = 5, a = {- 4, 9}, B = {0, - 4, 9}, then a ∩ B = {- 4, 9} and a ∩ B = {9} contradict
2 (56-x) - 10 = 30 + X (to solve the equation,
If the equation (2M2 + M-3) x + (m2-m) y-4m + 1 = 0 represents a straight line, then the real number m satisfies ()
A. m≠0B. m≠-32C. m≠1D. m≠1，m≠-32，m≠0
If equation (2M2 + M-3) x + (m2-m) y-4m + 1 = 0 represents a straight line, then 2M2 + M-3 and m2-m are not equal to 0 at the same time, and M = 1 is obtained from 2M2 + m − 3 = 0m2 − M = 0, so when m ≠ 1, 2M2 + M-3 and m2-m are not equal to 0 at the same time
Given two propositions, P: for any real number x, ax ^ 2 + ax ^ 2 + 1 > 0 is always true; Q: the equation x ^ 2-x + a = 0 about X has real roots
Given two propositions, P: for any real number x, ax ^ 2 + ax ^ 2 + 1 > 0 is always true; Q: the equation x ^ 2-x + a = 0 about X has real roots. If PVQ is true proposition and P \ / Q is false proposition, the value range of real number a is obtained
The proposition p must be wrong; for this proposition, we should think that the discriminant must be less than 0, right? For Q, the discriminant is greater than or equal to 0. Then we can see that P or Q has and only one false proposition (you have wrong number), and then we can discuss it according to the situation: 1. Q is false; 2. Q is true. Draw the number axis
3.5x = 3.56 + X to solve the equation,
3.5x=3.56+x ,
∴ 3.5x-x=3.56,
∴(3.5-1)x=3.56,
∴2.5x=3.56,
∴x=3.56÷2.5,
∴x=1.424
solution
3.5x=3.56+x
3.5x-x=3.56
2.5x=3.56
x=1.424
If you don't understand, please ask. If you understand, please take it as the best answer in time! (*^__ ^ *
3.5X-X=3.56
2.5X=3.56
X=3.56/2.5
X=1.424
3.5x=3.56+x
3.56x-x=3.56
2.5x=3.56
x=3.56÷2.5
x=1.424
424, thank you
If the equation (2m & sup2; + M-3) x + (M & sup2; - M) y-4m + 1 = 0 represents a straight line, then the real number m satisfies
A m ≠ 0, b m ≠ - 3 / 2, C M ≠ 1, d m ≠ 1, m ≠ - 3 / 2, m ≠ 0
C
The equation is transformed to (m-1) (2m + 3) x + m (m-1) y-4m + 1 = 0
The equation represents a straight line, then the coefficients of X and y cannot be zero at the same time
(m-1) (2m + 3) = 0 and m (m-1) = 0. When the coefficients of X and y are both zero, M = 1
So in order for the equation to represent a straight line, m must not be equal to 1, choose C
Answer: C
If this equation represents a straight line, then (2m & sup2; + M-3) and (M & sup2; - M) cannot be 0 at the same time
From 2m & sup2; + M-3 = 0
（2m+3)(m-1)=0
m=-3/2 m=1
From M & sup2; - M = 0
m=0 m=1
When m = 1, the equation cannot represent a straight line
(1 / 2) known proposition p: for any real number x, the square + ax + 4 > 0 of X is constant; proposition q: for the equation x, the square - 2x + a = 0 of X has a real root
(1 / 2) known proposition p: for any real number x, the square of X + ax + 4 > 0 is constant; proposition q: for the equation of X, the square of X - 2x + a = 0 has a real root. If P or q is true, P and Q are positive
First solve P, Q
P:a^2-16
How much is the solution to the equation x + 56-75 = 5x
x-19=5X
-19=4X
X=-19/4
If the focus of the hyperbola is (0,2), then the value of the real number m is
x^2/m-y^2/3m=1
The focus is on the y-axis
Then y ^ 2 is positive
So m
Isn't it good to substitute 0 and 2