Given the point m (3,2) n (0,1), the point P is on the x-axis, and the PM + PN is the shortest, then the point P coordinate is

Given the point m (3,2) n (0,1), the point P is on the x-axis, and the PM + PN is the shortest, then the point P coordinate is

On (1.5,0)
The equal sign a ^ + B ^ > = 2Ab holds only when a = B
So to minimize the sum of squares of the two segments, the two segments should be equal, so (3 + 0) / 2 = 1.5

For example: given a ^ 2-3a + 1 = 0, find the value of a ^ + 1 / A ^ 2. From a ^ 2-3a + 1 = 0, we know that a ≠ 0 ‖ A-3 + 1 / a = 0, that is, a + 1 / a = 3 ‖ a ^ 2 + 1 / A ^ = (a + 1 / a)
Example: given a ^ 2-3a + 1 = 0, find the value of a ^ + 1 / A ^ 2
A ≠ 0 from a ^ 2-3a + 1 = 0
A-3 + 1 / a = 0, i.e. a + 1 / a = 3
∴a^2+1/a^2=(a+1/a)²-2=7
Follow the example above,
Given y ^ + 3y-1 = 0, find the value of Y ^ 4 / y ^ 8-3y ^ 4 + 1

Given y ^ 2 + 3y-1 = 0, find the value of Y ^ 4 / y ^ 8-3y ^ 4 + 1
Know y ≠ 0 from y ^ 2 + 3y-1 = 0
That is, Y-1 / y = - 3
∴y^2+1/y^2=(y-1/y)^2+2=（-3）^2+2=11
If y ^ 4 / (y ^ 8-3y ^ 4 + 1) is divided by Y ^ 4, 1 / (y ^ 4 + 1 / y ^ 4-3) = 1 / [(y ^ 2 + 1 / y ^ 2) ^ 2-2-3] = 1 / (11 ^ 2-5) = 1 / 116

As shown in the figure, in the trapezoid ABCD, ad is parallel to CB, angle c is equal to 90 degrees, and AD + BC = AB AB AB is the diameter of circle O. prove that circle O is tangent to CD

Make a straight line parallel to AD and CB through the center O of circle O and intersect CD at E
O is the center of the circle, AB is the diameter, so o is the midpoint of AB; OE is parallel to CB, so e is also the midpoint of CD, so OE is the average of the upper and lower bottom of the trapezoid, OE = (AD + BC) / 2 = AB / 2, AB is the diameter of the circle O, OE is the radius of the circle O, e is a point on the circle;
Because angle c is a right angle and OE is parallel to BC, OE is perpendicular to CD. Therefore, circle O is tangent to CD. A line passing through the end of radius and perpendicular to radius is tangent to circle

If p1p2 = 2 (Q1 + Q2), it is proved that at least one of the equations x2 + P1x + Q1 = 0 and X2 + P2X + Q2 = 0 has real roots

Suppose that the original proposition is not tenable, that is, X2 + P1x + Q1 = 0 and X2 + P2X + Q2 = 0 ∧ △ 1 = p12-4q1 < 0, △ 2 = p22-4q2 < 0, the two formulas are added together to get: p12 + p22-4q1-4q2 < 0, that is, p12 + P22 < 4 (Q1 + Q2) and ∧ p1p2 = 2 (Q1 + Q2), ∧ p12 + P22 < 2p1p2p2, that is: (P1-P2) 2 < 0, which is obviously not tenable. Therefore, if the hypothesis is not tenable, the original proposition is correct

It is known that in rectangle ABCD, ab = 3cm, ad = 9cm, fold the rectangle so that point B coincides with point D, and the crease is EF, then the area of △ Abe is______ A．6cm2B．8cm2   C．10cm2D．12cm2．

Fold the rectangle so that the point B coincides with the point D, ∵ be = ed. ∵ ad = 9cm = AE + de = AE + be. ∵ be = 9-ae, according to the Pythagorean theorem, AB2 + AE2 = be2. The area of AE = 4. ∵ Abe is 3 × 4 ∵ 2 = 6

It is known that x = 3 is an extreme point of the function f (x) = AlN (1 + x) + x ^ 2-10x
(1) Finding a, (2) finding the monotone interval of F (x)

f'(x)=a/(1+x)+2x-10
x=3,f'(x)=0
a/4+6-10=0
a=16
f(x)=16ln(1+x)+x^2-10x
f'(x)=16/(1+x)+2x-10
=2(x-3)(x-1)/(x+1)
x> 3 or 11

The three vertices a (1,2), B (- 1, - 2), C (- 2,3) of △ ABC are translated to the point a '(- 1, - 2) so that a and a' coincide. Then the coordinates of B and C are______ ．

The abscissa of B after translation is - 1 + (- 1-1) = - 3; the ordinate is - 2 + (- 2-2) = - 6; the abscissa of C after translation is - 2 + (- 1-1) = - 4; the ordinate is 3 + (- 2-2) = - 1; so the answer is: (- 3, - 6), (- 4, - 2)

1. If the product of (x to the second power + PX + 3) (x to the second power - 2x + Q) does not contain the terms of X to the second power and X to the third power. (1) calculate the values of P and Q
(2) First simplify, then evaluate; (Q + 1) quadratic + (q-3) (Q + 3) + (q-3) (Q + 1) value

(x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\- 3) (Q + 3) = (- 1-3) (- 1 +...)

As shown in the figure, in the cube abcd-a1b1c1d1, e and P are the midpoint of edge BC and CC1 respectively. (1) verification: BD1 ‖ plane c1de; (2) verification: plane a1b1p ⊥ plane c1de

(1) Prove: as shown in Figure 1, connect CD1, intersect C1d at point O, ∵ e is the midpoint of BC, O is the midpoint of CD1, ∵ BD1 ⊄ plane c1de, OE ⊂ plane c1de, BD1 ⊂ plane c1de is known from the judgment theorem of line plane parallel. (2) prove that A1B1 ⊥ plane bcc1b1, C1E ⊂ plane bcc1b1, ≁ A1B1 ⊥ C1E, ⊥ b1c1o1 = ∠ CEc1, ⊂ c1o1 = ∠ cc1e, and b1c1 = C1C, thus RT △ b1c1p ≌ R T △ c1ce, ∩ c1p = CE, C1E ⊥ b1p. Also ∵ A1B1 ∩ b1p = B1, ∩ C1E ⊥ plane a1b1p. ∵ C1E ⊂ plane c1de, ∵ plane a1b1p ⊥ plane c1de

A simple operation of multiplying the fifth power of two by the fourth power of five

The fourth power of 2 multiplied by the fourth power of 5 is the fourth power of 10, equal to 10000, and then multiplied by the remaining 2, resulting in 20000