If f (x) is an even function defined on R and an increasing function on the interval from negative infinity to 0, and f (2x + 1) < f (3x + 2), the value range of X is obtained emergency

If f (x) is an even function defined on R and an increasing function on the interval from negative infinity to 0, and f (2x + 1) < f (3x + 2), the value range of X is obtained emergency

(-1,-2/3)

1. The normal vector of a plane. Can this vector pass through this plane?
Or a point on the plane and a point out of the plane, two vectors connecting the two points and perpendicular to the plane, is called normal vector?

The normal vector is a free vector perpendicular to the plane. It doesn't care about the position of the starting point, only the direction

As shown in the figure, △ ABC, CE is the extension of BC, CD bisects ∠ ace, CD / / AB, and proves that △ ABC is an isosceles triangle

Proof: because AB / / CD, so ∠ ABC = ∠ DCE ∠ ACD = ∠ ACB
And because CD bisects ∠ ace, ∠ ACD = ∠ DCE
Introduce ∠ BAC = ∠ ABC
It is proved that △ ABC is an isosceles triangle

The second car transports one car of coal. At the end of the transportation, the first car transports more than 25% of the total amount of 9 tons. The ton transported by the second car is equivalent to 50% of the coal transported by the first car. How many tons of coal does this pile have?

Suppose this pile of coal comes to x tons
25%x+9=x-（25%x+9）50%
3/2（1/4x+9）=x
3/8x+27/2=x
5/8x=27/2
x=27/2*8/5
=21.6 tons
This pile of coal is 21.6 tons

It is known that the center of circumcircle of triangle ABC is o. if the vector 3oa + 4ob + 5oC = 0, the angle c is obtained

Because the center of triangle ABC circumscribed circle is O, and 3oa + 4ob + 5oC = 0 (OA, ob, OC, 0 are vectors), then OA = (3oa + 4ob) / 5oC and OC dot product = | OC | ^ 2 = (3oa + 4ob) ^ 2 / 25 = 9 | ^ 2 / 25 + 24 (OA dot product OB) / 25 + 16 | ^ 2 / 25 = | OC | ^ 2 + 24 (OA dot product OB) / 25 = 9 | ^ 2 / 25

In △ ABC, ∠ BAC = 50 ° and ∠ B = 60 °, ad is the angular bisector of △ ABC, and the degree of ∠ ADC and ∠ ADB can be calculated,

∵∠BAC=50°,∠B=60°
∴∠C=180°-50°-60°=70°
∵ ad is the angular bisector of △ ABC
∴∠DAB=∠DAC=50°/2=25°
∴∠ADC=180°-25°-60°=95°
∴,∠ADB=180°-∠ADC=180°-95°=85°

How to use formula method to solve quadratic equation of one variable (normal format), 4Y square - 6 = 7Y

It's very simple

The center of the ellipse C is at the origin, the focus F1 and F2 are on the x-axis, a and B are the vertices of the ellipse C, pshi is a point on the ellipse C, and Pf1 ⊥ x-axis, Pf1 ∥ AB are used to calculate the eccentricity

(it should be PF2 / / AB) let the elliptic equation be x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, take the abscissa of P as - C, and substitute it into the ordinate of P as B ^ 2 / A, because PF2 / / AB, so the angle pf2f1 = angle ABO, the tangent value is B / A, so (b ^ 2 / a) / 2C = B / A, and change it into b = 2c, so a ^ 2-C ^ 2 = 4C ^ 2, a ^ 2 = 5C ^ 2, e ^ 2 = 1 / 5, E = √ 5 / 5

In △ ABC, ∠ BAC = 90 °, ad ⊥ BC, EF ⊥ BC, FM ⊥ AC, the vertical foot is D, F, m, ∠ 1 = ∠ 2 respectively, FM = FD is proved
Draw your own picture, don't pour water maliciously!

∠1=∠2,
Is it ∠ Abe = ∠ FBE?
If so, the solution is as follows
prove:
Connecting AF
In △ ABC, ∵ ∠ Abe = ∠ CBE, AE ⊥ AB, EF ⊥ BC,
∴AE=EF
∴∠EAF=∠EFA
Also ∵ ad ⊥ BC, EF ⊥ BC
∴AD‖EF
∴∠EFA=∠DAF
∴∠EAF=∠DAF
∵FM⊥AC
∴∠ADF=∠AMF=90°
In △ ADF and △ AMF, ∵ DAF = ∵ MAF, ∵ ADF = ∵ AMF, AF = AF
∴△ADF≌△AMF（AAS）
∴FD=FM

The quadratic equation with real coefficients x ^ 2 (1 + a) x + A + B + 1 = 0 has two real roots x1, X2 and 0
X ^ 2 + (1 + a) x + A + B + 1 = 0, the two real roots are x1, X2, and 0

x1>0,x2>0
x1x2>0
a+b+1>0
x1+x2>0,-(1+a)>0,a