Given that the solution of the equation x − 42-a + 1 = x is suitable for the inequalities - 12x ≤ - 1 and X-2 ≤ 0, the value of a is obtained

Given that the solution of the equation x − 42-a + 1 = x is suitable for the inequalities - 12x ≤ - 1 and X-2 ≤ 0, the value of a is obtained

The solution set of inequality-12x ≤ - 1 is x ≥ 2, the solution set of inequality X-2 ≤ 0 is x ≤ 2, so the value of X satisfying both inequalities is x = 2. Substituting x = 2 into x − 42-a + 1 = x, a = - 2 is obtained
Given that the minimum integer solution of inequality 5 (X-2) + 8 ＜ 6 (x-1) + 7 is the solution of equation 2x AX = 4, find the value of A
From 5 (X-2) + 8 ＜ 6 (x-1) + 7, we get x ＞ - 3, so the minimum integer solution is x = - 2. Substituting x = - 2 into 2x AX = 4, we get a = 4
It is known that y is the inverse proportional function of X, and when x is equal to - 6, y is equal to one-half, and the value range of the independent variable x is obtained
It is known that y is the inverse proportional function of X, and when x is equal to - 6, y is equal to one-half
The inverse scale function is y = - 3 / X
The value range of independent variable x {x | x ≠ 0}
It is known that the inverse proportion function y is equal to minus x, twelve y is equal to four-thirds of the independent variable x
Right?
Given the inverse scale function y = - 12 / x, when y ≤ 4 / 3, the value range of the independent variable x
-12/x≤4/3
x> 0, or X ≤ - 9
Under the following conditions, we can find out the value range of the quadratic function B ax + C, respectively
According to the image of the quadratic function y = AX2 + BX + C under the following conditions, the value ranges of a, B and C are obtained respectively according to the image of the quadratic function y = AX2 + BX + C under the following conditions: (1) about the y-axis symmetry. (2) the vertex of the function is on the y-axis. (3) the vertex has two intersections at the origin (4) and the y-axis, and is on both sides of the origin. The value ranges of a, B and C are obtained
1) On Y-axis symmetry
That is, the axis of symmetry is x = 0, so B = 0
2) The vertex of the function is on the Y axis
That is, the axis of symmetry is x = 0, so B = 0
3) Vertex at origin
It shows that y = ax ^ 2, so B = C = 0
4) There are two intersections with the X axis, and they are on both sides of the origin
If there are two roots, one positive and one negative, then there are two products = C / A0, that is, there are two different real numbers
So the condition is a, C different sign
(1) For Y-axis symmetry,
Then B = 0, a ≠ 0, C is any real number;
(2) The vertex of the function is on the Y axis
Then B = 0, a ≠ 0, C is any real number;
(3) Vertex at origin
Then B = 0, C = 0, a ≠ 0;
(4) There are two intersections with the Y axis, and they are on both sides of the origin.
Then AC < 0, a ≠ 0
(lgx) 2 + lgx ^ 3 + 2 = 0 to solve the equation
(logx)^2+lgx^3+2=0
(lgx)^2+lgx^3+2=0
(lgx)2+3lgx+2=0
(lgx+2)(lgx+1)=0
Lgx = - 2 or lgx = - 1
X = 1 / 100 or x = 1 / 10
(lgx)2+lgx^3+2=0
(lgx)2+3lgx+2=0
Let lgx = t, then
t²+3t+2=0
（t+1）（t+2）=0
So t = - 1 or T = - 2
That is lgx = - 1, so x = 0.1,
Or lgx = - 2, x = 0.01
(lgx)^2+lgx^3+2=0
(lgx)^2+3lgx+2=0
(lgx+1)(lgx+2)=0
Lgx = - 1, or lgx = - 2
X = 10 ^ (- 1) = 0.1, or x = 10 ^ (- 2) = 0.01
m. If n is opposite to each other and neither is zero, X and y are reciprocal to each other, find XY (M + n) - M / N + 2XY
This is the second term problem of grade 6, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16, p.16_ >
If M and N are opposite to each other and are not zero,
It is known that M + n = 0, M / N = - 1
The reciprocal of X and Y
We know that: xy = 1
therefore
xy（m+n）-m／n+2xy
=1×0-(-1)+2×1
=1+2
=3
m. N is opposite to each other, M + n = 0
X and y are reciprocal, xy = 1
xy（m+n）-m／n+2xy
=0+1+2
=3
Mathematical alphabet formula for Primary School Grades 1-6 (PEP)
Square C perimeter s area a side length perimeter = side length × 4 C = 4A area = side length × side length s = a × a 2 cube V: Volume A: edge length surface area = edge length × edge length × 6 s surface = a × a × 6 volume = edge length × edge length × edge length v = a × a × a 3 rectangle C perimeter s area a side length perimeter = (length + width) × 2
If negative a equals one third, how much is a? If negative a equals negative 7.7, how much is a? If 4x minus 5 and 3x minus 9 are opposite numbers, how much is x
A equals minus one-third a equals 7.7 x equals 2
If the image of quadratic function y = AX2 + BX + C passes through the points (- 1,0), (0,1), and the vertex is on the right side of y-axis, let s = a + B + C, then what is the value range of S?
∵ image passing through point (0,1) ∵ substituting x = 0, y = 1 into y = AX2 + BX + C to get: C = 1 ∵ image passing through point (- 1,0) ∵ substituting x = - 1, y = 0 into y = AX2 + BX + C to get: A-B + C = 0, that is, B = a + C = a + 1 ∵ s = a + B + C = a + (a + 1) + 1 = 2 (a + 1) you only need the value range of a now