Inequality 2x − The solution set of x > 1 is______

Inequality 2x − The solution set of x > 1 is______

Let t=
x（t≥0）
Then x = T2
Then inequality 2x −
X > 1 can be converted into
2t2-t-1＞0
T ＜ 1 is obtained
2, or T > 1
And ∵ t ≥ 0
∴t＞1
The solution is: x > 1
So the answer is: (1, + ∞)

0

(2x+5)/3≥x-5,
2x+5≥3x-15
-x≥-20
x≤20
When a = 2
(x+3)/2

It is known that the straight line L passes through the point P (3,4), and its inclination angle is twice that of the line root 3x-y + root 3 = 0? Find the L-line equation

Let the slope be K, then k = tan2x = 2tanx / (1-tanx * TaNx) = 6 / (1-9) = - 3 / 4
So the straight line is: y-4 = - 3 / 4 (x-3)
That is: 3x + 4y-25 = 0

The mathematical problem passes through the intersection point of the straight line: 2x + Y-3 = 0 and the straight line: 3x-2y-1 = 0, and the distance from the origin is root 2

Firstly, the coordinates of the intersection point are obtained, that is (1,1). Then, let the linear equation be y - 1 = k * (x - 1)
Finally, K value is obtained from the formula of distance from point to line

How many straight lines passing through the intersection of two straight lines X - (radical 3) y + 1 = 0 and (radical 3) x + Y - (radical 3) = 0, and whose distance from the origin is equal to 1? A ^ 2 + B ^ 2 under the formula d = | ax1 + bx2 + c1| / radical For detailed explanation

First find the intersection point X - √ 3Y + 1 = 0 (1)
√3x+y-√3=0 (2)
(2)*√3+(1) 4x-2=0 x=1/2
Substituting (2) y = √ 3 / 2
So the intersection point is (1 / 2, √ 3 / 2)
Let the linear equation be y = K (x-1 / 2) + √ 3 / 2, that is, kx-2y + √ 3-K = 0
The distance from the origin is known to be 1
Then the formula of distance from point to line d = I √ 3-KI / √ (k? 2 + 2?) = 1
Square of both sides 3-2 √ 3K + k? 2 = k? 2 + 4
The solution is k = - √ 3 / 6
So the linear equation is y = - √ 3x / 6 + √ 3 / 12 + √ 3 / 2
That is, y = (- √ 3 / 6) x + 7 √ 3 / 12
So only one line satisfies the condition

Given that the straight line L passes through the intersection point a of the lines X + y + 1 = 0 and 3x-y + 7 = 0, and the distance from the coordinate origin o is the root sign 5, the equation of the straight line L is solved

Let the equation y = K (x + 2) + 1 + 1 and the circle x + y = 5 have only one intersection point  5 = 5 = 5, x + y + 1 = 0,3x-y + 7 = 0 → x = - 2, y = 1 → a (- 2,1,1). Let the linear l equation y = K (x + 2) + 1 and the circle x + y = 5 have only a intersection point \\\\\\\\\\\\\\\\\\\\\\\(x + 2) + 1 = 2 (x + 2) + 1 = 2x + 5

(1) (2) given the parallel lines 3x + 2y-6 = 0 and 6x + 4Y = 0, find the locus of the point whose distance is equal to the two parallel lines

One
Let the slope K
Then the two parallel line equations are as follows:
y=kx ==> kx-y=0
y-3=k(x-1) ==> kx-y+(3-k)=0
Radical 5 = |3-k| / (k ^ 2 + 1) ^ (1 / 2)
5(k^2+1)=(3-k)^2
2k^2+3k-2=0
(2k-1)(k+2)=0
K = 1 / 2 or K = - 2
Therefore, two parallel line equations: y = (1 / 2) x and y = (1 / 2) x + (5 / 2)
Or: y = - 2x and y = - 2x + 5
Two
It is known that the parallel line 3x + 2y-6 = 0
And 6x + 4Y = 0 = = > 3x + 2Y = 0
The locus of a point equidistant from these two parallel lines:
3x+2y+[(-6+0)/2]=0
That is, 3x + 2y-3 = 0

Known (Radix 3x + 5y-2-a) + (Radix 2x + 3y-a) = (Radix x-2011 + y) * (Radix 2011-x-y)

For a hint, you can solve it yourself,
0+0=0,
Because the root must be positive or 0, x + y-2011 is greater than or equal to 0, and (x + y-2011) is greater than or equal to 0,
So x + y = 2011
That is (radical.) + (radix.) = 0,
That is, 0 + 0 = 0,3x + 5y-2-a = 0,2x + 3y-a = 0
We can solve x, y, a

The equation of the line L whose length is 3 times the root sign 2 is obtained by two straight lines l1:3x + 4y-7 = 0 and l2:3x + 4Y + 8 = 0 through point a (2,3)

Let the equation of the line l be Y-3 = K (X-2)
L:y=kx-2k+3...1
L1:3x+4y-7=0...2
L2:3x+4y+8=0...3
Replace Formula 1 with equation 2 to find the intersection point of L and L1
3x+4(kx-2k+3)-7=0
x=(8k-5)/(4k+3)...A
Then put a into formula 2
3A+4y-7=0
y=(k+9)/(4k+3)
So the intersection point of L and L1 is [(8k-5) / (4K + 3), (K + 9) / (4K + 3)]
Replace Formula 1 with equation 3 to find the intersection point of L and L2
3x+4(kx-2k+3)+8=0
x=(8k-20)/(4k+3)...B
Then substitute B into formula 3
3B+4y+8=0
y=(9-14k)/(4k+3)
So the intersection of L and L2 is [(8k-20) / (4K + 3), (9-14k) / (4K + 3)]
The length of the section is 3 √ 2, so the distance between the two intersections is 3 √ 2
3√2=√【[(8k-5)/(4k+3)-(8k-20)/(4k+3)]²+[(k+9)/(4k+3)-(9-14k)/(4k+3)]²】
63k²+432k-63=0
The solution is k = - 7 or K = 1 / 7
By substituting the value of K into L, two l equations are obtained
7x + y-17 = 0 or x-7y + 19 = 0
Another method:
The slope of the line L1 = - 3 / 4, and the slope of the line L2 = - 3 / 4, so L1 and L2 are two parallel lines. Their distance D is | 8 - (- 7) | / √ (3 | + 4?) = 3
The drawing shows that the angle between the straight line L and the two parallel lines is 45 degrees
Let the slope of the line l be K: 1
tan45°=|(k2-k1/(1+k1k2)|
Then 1 = | K - (- 3 / 4) | / (1-3k / 4)
The solution is k = - 7 or K = 1 / 7
Using the point oblique equation: Y-3 = K (X-2)
Then the value of K is substituted into the equation of L
7x + y-17 = 0 or x-7y + 19 = 0

The length of the line with 3 root sign 2 is obtained through the point P (2,3) and cut by two parallel lines 3x + 4y-7 = 0 and 3x + 4Y + 8 = 0 The cut length is 3 √ 2

First of all, the distance between the two parallel lines is 3D = | 8 + 7 | / root sign (3? + 4?) = 3. Then draw a graph to know that the angle between this line and two parallel lines is 45 ° and let the slope of the straight line be K, then | K - (- 3 / 4) / (1 - 3K / 4) = 1K = - 7, k = 1 / 7. Then we can get the equation of the line through a known point