The solution set of 1 x square. + 4 X-5 ＞ 0

The solution set of 1 x square. + 4 X-5 ＞ 0

(x+5)*(x-1)>0
x> 1 and X

Let X1 and X2 be the two roots of the quadratic equation x ^ 2 + x-3 = 0, then the value of X1 ^ 3-4x ^ 2 + 19 is
From the relationship between root and coefficient, we can get: X1 + x2 = - 1, from the known: X1 ^ 2 + x1-3 = 0, X2 ^ 2 + x2-3 = 0
This is X1 ^ 2 + x1-3 = 0, X2 ^ 2 + x2-3 = 0
How is it derived from X1 + x2 = - 1?

This is not from X1 + x2
It's from the known. If you look at it carefully, it's from the known
X1, X2 are the two roots of the quadratic equation x ^ 2 + x-3 = 0
So X1 ^ 2 + x1-3 = 0, X2 ^ 2 + x2-3 = 0

As shown in the figure, the parabola y = ax2-2x + 3 (a ≠ 0) intersects with the X axis at two points a and B, and intersects with the Y axis at point C, B (1, 0). (1) find the analytical formula of the parabola; (2) point P is the moving point on the line AB, passing P as PD ‖ AC, intersecting BC with D, connecting PC, when the area of △ PCD is the largest? If it exists, find out the coordinate of point Q; if not, explain the reason

(1) ∵ the parabola y = ax2-2x + 3 goes through B (1,0), ∵ 0 = A-2 + 3, ∵ a = - 1, that is, the analytical formula of the parabola is y = - x2-2x + 3; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp (3 points) (2) let P (m, 0), then Pb = 1-m. from (1), we can see that C (0, 3) a (- 3, 0), ∧ OC = 3 & nbsp; & nbsp; ab = 4, ∧ PD ∧ AC, ∧ PDB ∧ ACB, ∧ Deco = bpba, that is, DE3 = 1 − M4, ∧ de = 34 (1-m) (5) s △ PCD = s △ pbc-s △ PBD = 12pb · oc-12pb · De, = 12 (1-m) · 3-12 (1-m) · 34 (1-m), = - 38 (M + 1) 2 + 32, ∵ - 3 ≤ m ≤ 1, ∵ when m = - 1 & nbsp; & nbsp; s △ PCD has a maximum value of 32, ∵ P (- 1, 0) (8 points) ② there is a point Q on the line AC, so that △ PBQ is an isosceles triangle. The reasons are as follows: Method 1: ∵ P (- 1,0), B (1,0), ∵ Pb = 2, Op = ob, ∵ CP = CB, when QP = QB, ∵ Q coincides with C & nbsp; & nbsp; that is, q (0,3) (9 points) ∵ OA = OC = 3, ∵ △ OAC is an isosceles triangle, ∵ AB = 4 ∵ the distance between point B and straight line AC is ab · sin45 ° = 22, that is, BQ ≥ 22 ∵ BQ ≠ BP (11 points) when PQ = Pb = 2, PQ = PA, ∠ PQA = ∠ PAQ = 45 °, PQ ⊥ AB, q (- 1,2). In conclusion, there are points Q1 (0,3) and Q2 (- 1,2) such that △ PBQ is an isosceles triangle (13 points) method 2: ∵ P (- 1,0), B (1,0), ∵ Pb = 2, Op = ob, ∵ CP = CB, when QP = QB, ∵ Q coincides with C & nbsp; & nbsp; that is, q (0,3) (9 points) from a (- 3, 0) and C (0, 3), we can get the analytic formula of the straight line AC as y = x + 3, let Q (n, N + 3), and let Q be QF ⊥ X axis in F, then f (n, 0), ｜ pf = | - 1-N | = | n + 1 |, QF = | n + 3 | BF = | 1-N | = | n-1 |, ｜ bq2 = BF2 + QF2 = (n + 3) 2 + (n-1) 2 = 2 (n + 1) 2 + 8 & gt; 4, ｜ BQ ≠ BP (11 points) pq2 = PF2 + QF2 = (n + 1) 2 + (n + 3) 2 = 2n2 + 8N + 10, when PQ = Pb = 2, pq2 = 4, ｜ 2n2 + 8N + 10 = 4 & nbsp; & nbsp; the solution is n = - 1 or n = - 3 (12 points) ∵ n = - 3, Q coincides with a, P, B, q are on the same line, ∵ n = - 3 does not fit the problem, ∵ Q (- 1,2). To sum up, there are points Q1 (0,3), Q2 (- 1,2) such that △ PBQ is an isosceles triangle (13 points)

Simplify (a + B + C) (the square of a + the square of B + the square of C - AB BC CA) thank you

Open all the brackets according to the law of multiplication and distribution, and the result after merging the similar items is
a³+b³+c³-3abc

How can the parameter equation of curve C1 be transformed into a polar coordinate equation when x = 4 + 5cost y = 5 + 5sint

Firstly, it is transformed into rectangular coordinate equation: (x-4) / 5 = cost, (Y-5) / 5 = Sint)
=> (x-4)^2/5^2=cos^2t 、 (y-5)^2/5^2=sin^2 t
=> （x-4)^2/5^2+(y-5)^2/5^2=(cos t)^2+(sin t)^2=1
∴ （x-4)^2+(y-5)^2=5^2
Then it is transformed into polar equation: x ^ 2 + y ^ 2-8x-10y = - 16 = > ρ ^ 2-8 ρ OS θ - 10 ρ sin θ = - 16
=>ρ ^ 2-10 ρ sin θ - 8 ρ cos θ + 16 = 0

Some questions about factorization,
1. Square of X + X + M = (x-n) square, then M =, n=
2. Given 2x-y = 1 / 3, xy = 2, find the value of 2x ^ 4Y ^ 3-x ^ 3Y ^ 4
3. If 4x ^ 2-4x + 9y ^ 2-12y + 5 = 0, find the value of 6x + 2 / 3Y

1. Square of X + X + M = (x-n) square, then M =, n = x ^ 2 + X + M = x ^ 2-2nx + n ^ 2. Because of identity, there must be - 2n = 1, M = n ^ 2n = - 1 / 2m = n ^ 2 = 1 / 42. Given 2x-y = 1 / 3, xy = 2, find the value of 2x ^ 4Y ^ 3-x ^ 3Y ^ 4 = (XY) ^ 3 (2x-y) = 2 ^ 3 * 1 / 3 = 8 / 33

As shown in the figure, points a and B are two points on the unit circle, points a and B are in the first and second quadrants respectively, point C is the intersection of the circle and the positive half axis of X axis, and △ AOB is an equilateral triangle. If the coordinates of point a are (35, 45), mark ∠ COA = α
(1) Find the value of 1 + sin 2 α 1 + cos 2 α; (2) find the value of | BC | 2

(1) The coordinates of ∵ a are (35, 45). According to the definition of trigonometric function, sin α = 45, cos α = 35, ∵ 1 + sin2 α 1 + Cos2 α = 1 + 2Sin α, cos α 2cos2 α = 4918. (2) ∵ AOB is an equilateral triangle, ∵ AOB = 60 °. ∵ cos ∠ cob = cos (α + 60 °) = cos α cos60 ° - sin α

Limx →∞ (√ (x + 1) - √ x) limit

Molecular rationalization:
sqrt(x + 1) - sqrt(x) = 1 / (sqrt(x + 1) + sqrt(x)) -> 0

Quadratic function 1. Parabola y = 2x ^ - 4x + 1 can be obtained from parabola y = 2x ^ through the following translation: first, translate to? Units, and then to? Units
2. The axis of symmetry of the parabola y = one third of x ^ - 2x + 1 is? The fixed point coordinate is?
I don't want answers. I have answers. I don't know how to do it. Please tell me how to do it, that is, how to solve this kind of problem

1. First, the formula y = 2x ^ 2-4x + 1 is y = 2 (x-1) ^ 2-1,
So it's easy to see that it's y = 2x ^ 2
First right translation of a unit, and then down translation of a unit to get!
2. For the second question, you can also use the formula first, y = 1 / 3 (x-3) ^ 2-2,
Then the axis of symmetry is x-3 = 0, that is, the line x = 3, when x = 3, y = - 2, then the vertex coordinates are (3, - 2)
In fact, there is a formula y = a (X-H) ^ 2-k, vertex formula, the axis of symmetry is x = H,
The vertex coordinates are (h, K). You can directly insert them after you make the formula,
My answer is quite detailed,

A difficult problem (equation and arithmetic!)
After a period of time, the ratio of the distance between the express train and the whole journey is 2:3, and the distance between the local train and the local train is 180 km. The two trains continue to move forward at the original speed. When the express train reaches the local food, the local train only takes 6 / 7 of the whole journey

When the express arrives at yidishi, the local train only runs 6 / 7 of the whole journey, which means that the speed of the local train is 6 / 7 of that of the express train. So when the express runs 2 / 3 of the whole journey, the local train runs 2 / 3 * 6 / 7 = 4 / 7 and 1-4 / 7 = 3 / 7, which corresponds to 180 km, so the whole journey is 180 / (1-2 / 3 * 6 / 7) = 180 / (3 / 7) = 420 (km)