If the positive numbers a, B and C form an equal ratio sequence with a common ratio greater than 1, then when x > 1, logax, logbx, logcx () A. B. the reciprocal of each number is in turn in turn in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order

If the positive numbers a, B and C form an equal ratio sequence with a common ratio greater than 1, then when x > 1, logax, logbx, logcx () A. B. the reciprocal of each number is in turn in turn in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order in order

From the meaning of the question, we can get that the positive numbers a, B, C are not equal to 1, otherwise, at least one of the formulas in logax, logbx, logcx & nbsp; is meaningless. Because the positive numbers a, B, C form an equal ratio sequence with a common ratio greater than 1, then B2 = AC > 0, so when x > 1, there is logxb2 = logxac, that is, 2logxb = logxa + logxc, ‖ log

Given that a ＞ 0 and a ≠ 1, f (logax) = [A / (A & # 178; - 1)] × (x - (1 / x))
The expression of finding f (x)
Judging the parity and monotonicity of F (x)
When f (x) is (- 1,1), if f (1-m) + F (1-3m) < 0, the value range of M is obtained

(1) Let t = log (a) x, then x = a ^ t, so f (T) = [A / (A & # 178; - 1)] × [a ^ T-A ^ (- t)] that is, f (x) = [A / (A & # 178; - 1)] × [a ^ x-a ^ (- x)] (2) f (- x) = [A / (A & # 178; - 1)] × [a ^ (- x) - A ^ x] = - f (x), so that f (x) is an odd function

Given that a and B are normal numbers and not 1, find the value range of X that makes 4 (logax) ^ 2 + 3 (logbx) ^ 2 = 5 logax * logbx hold
Given that a and B are normal numbers and not 1, find the value range of x such that 4 (logax) ^ 2 + 3 (logbx) ^ 2 = 5 logax * logbx holds

4(logax)^2+3(logbx)^2=5logax*logbx
Divide the equation twice by (logbx) ^ 2
be
4（logax/logbx)^2+3=5logax/logbx
Then 4 (logab) ^ 2 + 3 = 5 logab
It has nothing to do with X, but only with a and B
So x > 0 is OK, but a and B must satisfy the above equation
However, the discriminant is less than 0
Therefore, the equation has no solution

cos20+cos40+cos140+cos100
It's all degrees

Cos[20 °] + Cos[40 °] + Cos[140 °] + Cos[100 °]
=Cos[20 °] - Sin[10 °]

The perimeter of the bottom of a cuboid is a square of 4 square centimeters, and its side view is also a square. Find the volume of the cuboid

Height = base circumference = 4cm
Bottom length = 4 / 4 = 1cm
Volume = 1 * 1 * 4 = 4 cc

A cuboid with a perimeter of 8 decimeters on the bottom and a square on the side. What is the cuboid's volume

32 cubic decimeter

The length of the cuboid is 12 cm and the height is 8 cm. The sum of the areas of the two sides of the shadow is 200 square cm
The shadow is on the left and right

The left and right sides are surrounded by the width and height of the cuboid,
Area of each surface = 200 △ 2 = 100 (square centimeter)
Width of cuboid = 100 △ 8 = 12.5 (CM)
Volume of cuboid
=12×12.5×8
=1200 (cm3)

As shown in the figure, it is a cuboid. The total area of the shadow part is 180 square centimeters. What is the cuboid's volume in cubic centimeters?

Let the width of the cuboid be a, then 12a + 8A = 180, & nbsp; & nbsp; & nbsp; & nbsp; 20A = 180, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; a = 9, 12 × 9 × 8 = 864 (cubic centimeter); a: the cuboid's volume is 864 cubic centimeter

A cuboid is 8 cm in length, 6 cm in width and 5 cm in height. Its bottom area is______ Square centimeter, the surface area is______ Square centimeter, the volume is______ Cubic centimeter

8 × 6 = 48 (square centimeter), (8 × 6 + 8 × 5 + 6 × 5) × 2, = (48 + 40 + 30) × 2, = 118 × 2, = 236 (square centimeter), 8 × 6 × 5 = 240 (cubic centimeter), answer: its bottom area is 48 square centimeter, surface area is 236 square centimeter, volume is 240 cubic centimeter. So the answer is: 48; 236; 240

There is a cuboid with a surface area of 184 square centimeters, a bottom area of 20 square centimeters, and a bottom perimeter of 18 centimeters. The cuboid has a volume of______ Cubic centimeter

20 × [(184-20 × 2) △ 18], = 20 × [(184-40) △ 18], = 20 × [144 △ 18], = 20 × 8, = 160 (cubic centimeter). Answer: the volume of this cuboid is 160 cubic centimeter. So the answer is: 160