It is proved that the + equation LG (2x) * LG (3x) = 1 has two equal real roots, and the product of the solutions of the two real roots is obtained

It is proved that the + equation LG (2x) * LG (3x) = 1 has two equal real roots, and the product of the solutions of the two real roots is obtained

LG (2x) * LG (3x) = 1 ℅ (LG2 + lgx) * (Lg3 + lgx) = 1, let t = lgx  T & # 178; + (LG2 + Lg3) t + LG2 * lg3-1 = 0 (* *) discriminant = (LG2 + Lg3) &# 178; - 4 (lg2lg3-1) = (lg2-lg3) &# 178; + 4 > 0 hold constant. There are two unequal roots T1, T2, and the original equation also has two unequal roots

F (x) = sin ^ 2 θ SiNx + cos ^ 2 θ cosx, who can help me simplify?
The one above is the square, not "2 θ", but "sin θ sin θ"

According to sin ^ 2 θ = (1-cos2 θ) / 2 cos ^ 2 θ = (Cos2 θ - 1) / 2
2 / 2 root sign 2 × sin (x - π / 4) (1-cos2 θ) = root sign 2 * sin (x - π / 4) sin ^ 2 θ

If f (SiN x / 2) = cosx + 1, then f (COS X / 2) =?

f(sinx/2)=cosx+1 =1-2(sin(x/2))^2 +1 =2- 2(sin(x/2))^2f(x)=2-2x^2f(cos(x/2)^2=2-2(cos(x/2))^2 =2-(1+cosx) =1-cosx

·Let the two focal points of the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1 be F1 and F2 respectively; p be a point on the ellipse, and find the abscissa range of P such that the angle f1pf2 is an obtuse angle
Parametric equation
The answer is (- 5 / 4 radical 7,5 / 4 radical 7)

Use the geometric method to make a circle with F 1 and F 2 as the diameter. The equation of a circle can be obtained from the equation of an ellipse. The two equations can be combined to get the solution x 1 and x 2 between them. Because we know that any point on the circle and the line connecting the diameter form a right angle. The line connecting the point of a circle and the diameter can form an obtuse angle

If the angle f1pf2 is 60 degrees, find the area of the triangle f1pf2
If the angle f1pf2 is 60 degrees, calculate the area of the triangle f1pf2 and the coordinates of P,

|PF1|+|PF2|=2a
|PF1|^2+|PF2|^2+2|PF1||PF2|=100
PF1|^2+|PF2|^2-2|PF1||PF2|cos60°=64
Subtraction: | Pf1 | PF2 | = 12
1/2|PF1||PF2|sin60°=3√3
1/2*F1F2*y=3√3
y=3√3/4
Then the abscissa can be obtained by introducing the equation

P on the ellipse x square / 4 + y square / 5 = 1, F1 and F2 are two focal points, and the angle f1pf2 = 30 degrees, the area of triangle f1f2p is calculated

In △ f1f2p, we get F1F2 & # 178; = Pf1 & # 178; + PF2 & # 178; - 2 * Pf1 * PF2 * cos30 ° from diagonal chord theorem, that is, F1F2 & # 178; = (Pf1 + PF2) &# 178; - 2 * Pf1 * PF2 * cos30 ° from above formula

F1, F2 are the focus of the ellipse x squared / 9 + y squared / 5 = 1, P is the point on the ellipse and f1xf2 = 15 to find the angle f1pf2

∵x^2/9+y^2/5=1
∴a^2=9,b^2=5
∴c^2=a^2-b^2
=4
c=±2
The coordinates of focus F1 and F2 are (- 2,0) and (- 2,0) respectively
What does f1xf2 = 15 mean?

Given that the hyperbola is 16x square-9y square = 144, find the focal coordinates, vertex coordinates, real axis length, imaginary axis length and eccentricity

First, write the hyperbola as the standard form
(x^2)/9-(y^2)/16=1
Half focus = root 16 + 9 = 5
Focus coordinates (5,0), (- 5,0)
Real axis length = 2 * radical 9 = 6
Imaginary axis length = 2 * radical 16 = 8
Vertex (3,0), (- 3,0), (0,4), (0, - 4)
Eccentricity e = 5 / 3

The process of finding the coordinates of the long focus and vertex of the major axis and minor axis of the ellipse x ^ 2 + 9y ^ 2 = 25

x^2+9y^2=25
☞：x²/25+y²/(25/9)=25/9
Then a & sup2; = 25, B & sup2; = 25 / 9
Long axis length 2A = 10, short axis length 2B = 10 / 3
Vertex (- 5,0), (5,0), (0,5 / 3), (0, - 5 / 3)
Signature: Ning Ying

P is the point on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1, F 1 and F 2 are the left and right focus, if the angle f 1pf 2 = 60 degrees, find the value of | Pf1 | PF2 |

F1 and F2 are (- 3,0), (3,0)
∠F1PF2=60°
PF1+PF2=2a=10
(F1F2)^2=(PF1)^2+(PF2)^2-2PF1*PF2*cos60°
36=(PF1+PF2)^2-3PF1*PF2
36=10^2-3PF1*PF2
|PF1||pF2|=64 /3