If the solution set of the inequality system x-a > = 1 2x-6 > 0 is x > 3, then the value range of a

If the solution set of the inequality system x-a > = 1 2x-6 > 0 is x > 3, then the value range of a

X-a > = 1 to get x > = a + 1
2x-6 > 0 results in x > 3
The solution set is x > 3
So a + 1
a≤2
It should be
x> = a + 1 and x > 3
The number axis method can get a + 1
If the system of inequalities {x2x-1} has no solution, find the range of K
I'm sorry
wrote it wrong
It should be {x2k-1
There is a solution when k = 0
K+1
K>2
Negative infinity to zero
Negative infinity to zero
It's impossible. There must be a solution. Is it wrong
X
Inequality KX ^ 2-2x + 1-k
In F (k) = K (x ^ 2-1) - 2x + 1, where x is regarded as a constant, the variable is K
It is a monotonic function
Just like a function f (x) = (a ^ 2-1) x-2a + 1, a is regarded as a constant, and the variable is X
If the complete set is a real number set R and the set a = {x | Log1 / 2 (2x-1) > 0}, what is the complement of a in R
2x-1 > 0, x > 0.5, the complement is X
Solve the equations 7y-x = 4,2x + y = 3
7y-x=4①,2x+y=3②
② X 7, 14x + 7Y = 21
③ - 1, 15x = 17
The solution is x = 17 / 15
Substituting x = 17 / 15 into the existing one, we get 2 × 17 / 15 + y = 3
The solution is y = 11 / 15
The solutions of the original equations are x = 17 / 15, y = 11 / 15
14y-2x=8
+The second formula is 15y = 11
y=11/15
x=17/15
Y = 7x-4
Take x = 7y-4 into 2x + y = 3 to get 14y-8 + y = 3 and simplify to 15y = 11
....
7y-x=4 （1）
2x+y=3 （2）
From (1) * 2 + (2)
2（7y-x)+(2x+y)=2*4+3
It is reduced to 15y = 11
Y = 11 / 15
Substituting (2)
We get 2x + 11 / 15 = 3
The result is x = 17 / 15
x=17/15 y=11/15
7y-x=4-----------(1)
2x+y=3-----------(2)
(2) * 7
7y+14x=21--------(3)
(3) - (1) get
15x=17
x=17/15
Substituting x = 17 / 15 into (1) yields
y=11/15
x=25/13,y=11/15
14y-2x=8
2X + y = 3, that is, 15y = 11, y = 11 / 15
Similarly, x = 25 / 13
On the positional relationship between circles
The radius wind of circle O1 and O2 is 7 and 2 respectively, the circle distance is 13, the line AB tangents circle O1 at point a, and tangents circle O2 at point B, and finds the length of AB?
Common tangent, two circles are separated, the inner and outer values are 13 ^ 2 - (7-2) ^ 2 under the root sign and 13 ^ 2 - (7 + 2) ^ 2 under the root sign
There are three situations that need to be discussed.
1.12 2.13 3. Radical 194
There are 1 column monomials: - 2x ^ 2, - 3x ^ 3,4x ^ 4
(1) Write the 99th and 100th monomials according to the arrangement rules of Ju's listed monomials;
(2) Write the expression of the nth (n is an integer) monomial
99 items - 100x ^ 100
100 items 101x ^ 101
Nth - (n + 1) (- x) ^ (n + 1)
Decomposition factor: 2x2 + xy-y2-4x + 5y-6
2x2+xy-y2-4x+5y-6=（x+y）（2x-y）-4x+5y-6 =（x+y）（2x-y）+2（x+y）-6x+3y-6 =（x+y）（2x-y+2）-3（2x-y+2）=（2x-y+2）（x+y-3）．
If the real numbers x and y satisfy the radical X-2 + (y + 1) &# 178; = 0, then X-Y is equal to
√x-2 + (y+1)²=0
Then X-2 = 0, y + 1 = 0
x=2 y=-1 x-y=3
Radical X-2 + (y + 1) ² = 0
Then X-2 = 0, x = 2
y+1=0, y=-1
x+y=1
If √ [X-2 + (y + 1) ²] = 0, then: X-2 = 0 ①, y + 1 = 0 ②. ① Then: X-Y = 3.
Three X-2 = five y + 4 2x-7y = 90 to solve the equations
x=-4,y=-14
Take 15 on both sides
5x-10=3y+12
5x-3y=22 (1)
2x-7y=90 (2)
(1)×7-(2)×3
35x-6x=154-180
29x=-26
therefore
x=-26/29
y=(5x-22)/3=-256/29
What you write in this equation is ambiguous: X / 3 - 2 = Y / 5 + 4
Because 2x-7y = 90; deformation: x = (7Y + 90) / 2; into the equation: (7Y + 90) / 6 - 2 = Y / 5 + 4
Both sides at the same time * 30 variant: 35y + 450-60 = 6y + 120 get y = - (270 / 29) calculate X by yourself
（x-2）/3=(Y+4)/5
In the same way, y = - 14, x = - 4
Hope to adopt!