Given that the equations x-by = 1, y-ax = 1, BX + ay = 1 have solutions, try to explain a * a + b * a + AB + A + B = 1

Given that the equations x-by = 1, y-ax = 1, BX + ay = 1 have solutions, try to explain a * a + b * a + AB + A + B = 1

X-by = 1, (1) y-ax = 1 (2) solving equation: (1) * a + (2): ax aby + y-ax = a + 1y = (1 + a) / (1-ab), substituting (1), x = (1 + b) / (1-ab); substituting: BX + ay = 1a (1 + a) / (1-ab) + B (1 + b) / (1-ab) = 1A + A ^ 2 + B + B ^ 2 = 1-aba ^ 2 + B ^ 2 + AB + A + B = 1
If a < B, x < y, compare ax + by with BX + ay
ax+by-bx-ay=(a-b)(x-y)>0
ax+by>bx+ay
It is known that "there exists x ∈ R, x ^ 2 + 2aX + 1"
"There exists x ∈ R, x ^ 2 + 2aX + 1
Discriminant = 4A ^ 2-4 > 0 a ^ 2 > 1
a> 1 or A1 or a
The equation of symmetry axis of image with function y = cos (2x + 2 / 7 π)
It's where COS is the highest
So cos (2x + 2 / 7 π) = ± 1
2x+2π/7=kπ
So x = k π / 2 - π / 7
2x+2/7π=kπ
x=-π/7+kπ/2
For any x ∈ R, there exists m ∈ r such that 4 ^ X-2 ^ (x + 1) + M = 0
4^x-2^（x+1）+m=0
(2^x)^2-2*2^x+m=0
If the proposition is not p, it is a false proposition
Then the proposition p is true
And T = 2 ^ x ＞ 0
So for any t > 0, there exists m ∈ r such that T ^ 2-2t + M = 0
Let f (T) = T ^ 2-2t + m, then f (1) = 1 ^ 2-2 + M = M-1 ≤ 0
So m ≤ 1
If you don't understand, please hi me, I wish you a happy study!
If proposition p is not a false proposition, then proposition p is a true proposition. That is to say, for any x ∈ R, m ∈ R can always be found so that the equation has real roots.
4^x-2^(x+1)+m=0
(2^x)²-2×2^x+m=0
(2^x-1)²=1-m
1-m≥0
m≤1
A symmetry axis equation of the image with the function y = cos [2x + (5 π) / 2] is?
The answer is that x equals π / 4
Y = cos (2x + 5 π / 2) = cos (2x + π / 2) = - sin2x, its axis of symmetry is a straight line passing through the highest or lowest point and perpendicular to the X axis,
So let 2x = π / 2 + K π, the axis of symmetry is x = π / 4 + K π / 2, K is an integer,
Taking k = 0, we obtain an axis of symmetry x = π / 4
First of all, you need to know that the image of y = cos X has two axes of symmetry in a cycle, that is, a straight line parallel to the Y axis passing through the highest and lowest points of the image. Then, y = cos [2x + (5 π) / 2], that is, y = cos [2 (x + π / 4)]. A very simple method is to draw a cycle of this function, and the result will come out naturally!
y = sinx μ???3??á?a x = (k + 1/2)|D y=sin??2x+￡¨5|D￡?/2????3??á ?a 2x + (5|D)/2 = (k + 1/2)|D 2x = (k -2)|D x =
If there is a real number such that | x-a | + | X-1|=
|X - a | can be regarded as the distance from the number axis to the point represented by A
|X - 1 | can be regarded as the distance from the number axis to the point represented by 1
Make the sum of the two distances less than or equal to 3
That is to say, the sum of two line segments does not exceed 3
Take exactly 3, a = 4 or a = - 2
The existence of real number makes the inequality hold
When - 2 ≤ a ≤ 4, there are real numbers that make the inequality hold
When a does not take this range, there is no real number to make the inequality hold
So the value range of a is - 2 ≤ a ≤ 4
What is the equation of symmetry axis of image with function y = cos (2x + π / 2)?
（2k-1）π/4
Given the function f (x) = − x2 + ax, X ≤ 1ax − 1, x > 1, if ∃ x1, X2 ∈ R, x1 ≠ X2, f (x1) = f (x2) holds, then the value range of real number a is ()
A. A ＜ 2B. A ＞ 2C. - 2 ＜ a ＜ 2D. A ＞ 2 or a ＜ - 2
Given that the image of quadratic function is x = 2 through (1,4) and (5,0), try to find the expression of quadratic function
It can be seen from the symmetry axis X = 2 that the image intersects the X axis at (5,0), (- 1,0)
So there are
Y = K (X-2) + B, and then (5,0), (- 1,0) are substituted,
Then we can get: 0 = K (5-2) + B
0=k(-1-2)+b
The solution is K and B
f(x)=ax^2+bx+c
∵ the axis of symmetry is x = 2
Let f (x) = a (X-2) ^ 2 + B
∵ through (1,4) and (5,0)
∴a+b=4;9a+b=0
∴a=-1/2;b=9/2
To substitute: F (x) = (- 1 / 2) (X-2) ^ 2 + 9 / 2 = (- 1 / 2) x ^ 2 + 2x + 5 / 2