The solution inequality of 3 / 5x-1 minus x > 1

The solution inequality of 3 / 5x-1 minus x > 1

(5x-1)/3-x>1
(5x-1)-3x>3
2x>4
X>2
The equation of the line L which is cut by the lines L1: 4x + 8y + 18 = 0 and L2: 2x + 4y-1 = 0 and whose length is root 10 and passes through point a (4,18)
∵ the slope of line L1 is K1 = - 4 / 8 = - 1 / 2, and the slope of line L2 is K2 = - 2 / 4 = - 1 / 2
∴L1∥L2
Then the distance between L1 and L2 d = | 9 - (- 1) | / √ (4 + 16) = √ 5
The length of the line cut by L1 and L2 is √ 10
The angle between L and L1 and L2 is 45 degrees
From the formula of the angle between two straight lines, Tan 45 ° = | (K + 1 / 2) / (1-k / 2)|
The solution is k = - 3 or 1 / 3
Therefore, the linear equation with slope k of point a (4,18) is y-4 = - 3 (x-18) or y-4 = (1 / 3) (x-18)
That is, 3x + y-58 = 0 or x-3y-6 = 0
Let the slope of the straight line l be K, it is easy to know that the distance between L1 and L2 is d = root sign 5, and the length of the line segment L is cut by L1 and L2 is root sign 10, so the angle between L and L1 (or L2) is 45 °. From the angle formula of two straight lines, | (K + 1 / 2) / (1-k / 2) | = 1, the solution is k = - 3
So the linear equation passing through point a (4,18) is y-4 = - 3 (x-18) or y-4 = (1 / 3) (x-18)
3x + y-58 = 0 or x-3y-6 = 0
Let the slope of the straight line l be K, it is easy to know that the distance between L1 and L2 is d = root sign 5, and the length of the line segment L is cut by L1 and L2 is root sign 10, so the angle between L and L1 (or L2) is 45 °. From the angle formula of two straight lines, | (K + 1 / 2) / (1-k / 2) | = 1, the solution is k = - 3
So the linear equation passing through point a (4,18) is y-4 = - 3 (x-18) or y-4 = (1 / 3) (x-18)
3x + y-58 = 0 or x-3y-6 = 0
The solution of inequality 2x-1 > half x is []
2x-1>1/2 x
3/2x>1
x>2/3
(2-1/2)x>1
x>2/3
The positional relationship between circle C1: x2 + Y2 + 4x + 8y-5 = 0 and circle C2: x2 + Y2 + 4x + 4y-1 = 0 ()
A. Intersect B. circumscribe C. inscribe D. circumscribe
It is known that circle C1: x2 + Y2 + 4x + 8y-5 = 0 is: (x + 2) 2 + (y + 4) 2 = 25; circle C2: x2 + Y2 + 4x + 4y-1 = 0: (x + 2) 2 + (y + 2) 2 = 9, then circle C1 (- 2, - 4), C2 (- 2, - 2), the distance between the centers of two circles C1C2 = (− 2 + 2) 2 + (− 2 + 4) 2 = 2 = 5-3, which is equal to the difference of radius, so the two circles are inscribed, so C
Solve the equation: negative one third (6-5x) = 8, 2x-1 minus 10x + 1 = 2x + 1, 2x + 1, x + 2, x-4 = 2
solve equations
Negative one third (6-5x) = 8
6-5x=-24
-5x=-30
X=6
2x-1 / 3 minus 10x + 1 / 6 = 2x + 1 / 4
4(2x-1)-2(10x+1)=3(2x+1)
8x-4-20x-2=6x+3
-18x=-9
x=0.5
One third x + one half (two-thirds x-4) = 2
x/3+x/3-2=2
2x/3=-4
x=-6
1. It is known that y = - 1 is the solution of equation 2Y's cube-2y's square + 4Y + m-5 = 0, find the square of M + M-1
2. If the absolute value of (M-3) x multiplied by (2m-5) - 4m = 0 is a one variable linear equation about X, find the square of M - 2m + 1
Y = - 1 is the solution of equation 2Y's Cube - 2Y's square + 4Y + m-5 = 0
2*(-1)^3-2*(-1)^2+4*(-1)+m-5=0
-2-2-4+m-5=0
m=13
Square of M + M-1
=13^2+13-1
=169+12
=182
On the inequality of X, if the absolute value of the sum of 2 / 3x plus m is less than 1, then the solution set of X is greater than 2 / 3 and less than 2, then the value of M is____ ?
To solve the inequality given by the problem, we obtain - 1 ≤ 3x / 2 + m ≤ 1, that is, 2 / 3 (- 1-m) ≤ x ≤ 2 / 3 (1-m). Therefore, 2 / 3 (- 1-m) = 2 / 3 and 2 / 3 (1-m) = 2 hold at the same time
How to calculate the square of 4Y = 2Y?
4y^2=2y
4y^2-2y=0
2y(2y-1)=0
So y = 0 or y = 1 / 2
4Y & sup2; = = 2Y, 4Y = 2, y = half, y = 0 also holds
(4Y) & sup2; = 2Y to get (4Y) & sup2; - 2Y = 0 to get 2Y (8y-1) = 0 to get y = 0 or y = 1 / 8
Solving inequality 2x-5 of 6 ≤ 3x of 4 + 2 of 1-3
2x-5 of 6 ≤ 3x of 4 + 2 of 1-3
2(2x-5)≤3(3x+1)-8
4x-10≤9x+3-8
4x-9x≤-8+13
-5x≤5
x≥-1
2(2X-5)≤3(3X+1)-8
4X-10≤9X+3-8
-5X≤5
X≥-1
How to calculate (2Y + 1) (2y-1) (the square of 4Y + 1)?
(2Y + 1) (2y-1) (square of 4Y + 1)
=(4y²-1)(4y²+1)
=The fourth power of 16y-1
(2y+1)(2y-1)(4y^2+1)=(4y^2-1)(4y^2+1)=16y^4-1;
(2Y + 1) (2y-1) = 4Y ^ 2-1 is the square difference formula (a + b) (a-b) = a ^ 2-B ^ 2
What is the topic? After reduction, it is 16 ^ 4-1