If the inequality | 2A − 1 | is constant for all non-zero real numbers x, then the value range of real number a is______ ．

If the inequality | 2A − 1 | is constant for all non-zero real numbers x, then the value range of real number a is______ ．

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A trapezoid, whose bottom is shortened by 4.5 meters, becomes a square, and its area is reduced by 18 square meters. How many square meters is the original trapezoid area
Please do it in the way of grade five!

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The height and upper bottom of trapezoid are as follows:
18 × 2 △ 4.5 = 8 (m)
The original trapezoid area is:
8 × 8 + 18 = 82 (M2)

Let f (x) = the absolute value of lgx, a and B be real numbers satisfying f (a) = f (b) = 2F [(a + b) / 2], where 0

The snail climbs up along the shaft wall at the bottom of the well. It climbs up 3m in the daytime and retreats 2m at night. A well is 12m deep. It climbs up from the bottom of the well and reaches the wellhead in a few days

The last day did not slide, before climbing
12-3 = 9 (m)
Yes
9 (3-2) = 9 (days)
Climb to the wellhead
9 + 1 = 10 (days)

If the real numbers a, B, C satisfy a + B + C = 0, ABC = 1, it is proved that at least one of a, B, C is not less than 2 / 3

Because ABC = 1, so C = 1 / AB, substitute C into a + B + C = 0 to get a + B + 1 / AB = 0, and multiply both sides by a to get a ^ 2 + Ba + 1 / b = 0. From the meaning of the question, we know that a, B, C satisfy a + B + C = 0; therefore, a, B must also satisfy a

A triangle and a parallelogram have the same area and height. If the bottom of the triangle is 10 cm, then the bottom of the parallelogram is 10 cm______ Cm

A: the bottom of a parallelogram is 5cm. So the answer is: 5

It is known that P is a moving point on the parabola y = 1 / 2x & sup2;. If the projection of point P on the X axis is m and the coordinates of point a are (6,17 / 2), then the minimum value of PA + PM is?

For the convenience of writing, remove the absolute value sign: (this problem does not need to find the value of P0, for the convenience of understanding, so find out)
Y = 1 / 2x & amp; sup2; & nbsp; focus f (0,0.5), collimator y = - 0.5 & nbsp;, extend PM collimator to H point, then & nbsp; PA = pH
PM=PH-0.5=PA-0.5
PM + PA = pf + pa-0.5, we can only find the minimum value of & nbsp; PF + PA & nbsp
From the fact that both sides of the triangle are longer than the third, PF + PA & gt; = & nbsp; fa & nbsp; (straight line segment), (formula 1)
Let the straight line FA and the parabola intersect at point P 0, then p 0 & nbsp; (3,4.5) can be calculated, and the other intersection (- 1 / 3,1 / 18) can be rounded off
When p coincides with P0, (formula 1) can get the minimum value, and FA = 10
The result is & nbsp; PM + PA = 9.5 & nbsp;
Note: the coordinate of point a in the figure is wrongly written, which should be (6,17 / 2)

(a ^ 2 + 8A ^ 2) + 22 (a ^ 2 + 8a) + 120 factorization

(a^2+8a)^2+22(a^2+8a)+120
=(a^2+8a+10)(a^2+8a+12)
=(a^2+8a+10)(a+2)(a+6)
The title is wrong, it has been changed

It is known that AB is the diameter of circle O, ad is perpendicular to CD, AC bisector angle DAB, and point C is on circle O. (1) prove that the straight line CD is the tangent of circle o

AC bisecting angle DAB, we can get the angle DAC = angle cab
Because angle ACB = 90 degrees, angle CBA + angle cab = 90 degrees
We can know from the above two formulas: angle CBA + angle DAC = 90
Because ad is perpendicular to CD, angle DCA + angle DAC = 90
So angle CBA = angle DCA
Because the angles CBA and DCA are chord angles, the theorem of chord angle deduces that CD is the tangent of circle o

No matter what the real values of M and N are, if the line (3m – n) x + (M + 2n) y – n = 0 passes through the fixed point P, then the coordinates of point P are

Solution 1: from (3m – n) x + (M + 2n) y – n = 0, m (3x + y) + n (- x + 2y-1) = 0, let 3x + y = 0, - x + 2y-1 = 0, the solution is: x = - 1 / 7, y = 3 / 7, so the coordinate of P point is (- 1 / 7,3 / 7). When x = - 1 / 7, y = 3 / 7 is substituted into the straight line (3m – n) x + (M + 2n) y – n = 0, so p (- 1 / 7,3 / 7) is the fixed point