Use the formula containing letters to express the difference between X and 5 times of 2.6 () the quotient of a divided by B () 50 minus 3 times of X () 5 times of a plus 7.5 ()

Use the formula containing letters to express the difference between X and 5 times of 2.6 () the quotient of a divided by B () 50 minus 3 times of X () 5 times of a plus 7.5 ()

|x-2.6*5|
b/a
|50-x|*3
5a+7.5

If f (x) = x & sup2; - a ㏑ x increases on (1,2], G (x) = x-a √ x decreases on (0,1), find the expression of F (x), G (x)
It is proved that f (x) - G (x) = x & sup2; - 2x + 3 has a unique solution when x > 0

∵ f (x) = x & sup2; - a ㏑ x increases on (1,2],
On (1,2], f '(x) = 2x-a / x = (2x & sup2; - a) / X is always greater than 0
Ψ 2x & sup2; - A is always greater than 0 on (1,2]
The minimum value of a ≤ (2x & sup2;) on (1,2]
That is, a ≤ 2
Similarly, ∵ g (x) = x-a √ x decreases on (0,1)
In (0,1), G '(x) = 1-A / (2 √ x) = [(2 √ x) - A] / (2 √ x) is always less than 0
(2 √ x) - A is always less than 0 on (0,1)
The maximum value of a ≥ (2 √ x) on (0,1)
That is, a ≥ 2
In order to satisfy a ≥ 2 and a ≤ 2 at the same time, it can only be a = 2
∴f(x)=x²-2㏑x,g(x)=x-2√x
Second question:
Let H (x) = f (x) - G (x) - (X & sup2; - 2x + 3) = x-2lnx + 2 √ x-3
Then the problem can be transformed into proving that h (x) = 0 has a unique solution when x > 0
The derivation of H (x) gives H '(x) = 1-2 / x + 1 / √ x = (√ x + 2) (√ x-1) / X
When x > 1, H '(x) > 0, H (x) increases
When x

Can 1 / 6 * 2 / 3 / (4 / 5-8 / 15) be simplified

1 / 6 * 2 / 3 / (4 / 5-8 / 15)
=1 / 9 / (4 / 15)
=5 out of 12

Given sin (π / 2 + α) = 12 / 13 (- 12 / 13 < α < 0), find cos (α - π / 6)

Sin (π / 2 + α) = cos (α) = 12 / 13, so sin (α) = 5 / 13:;
Cos (α - π / 6) = cos (α) * cos (π / 6) - sin (α) * sin (π / 6) = (12 (radical 3) - 5) / 26

As shown in the figure, the area of the shadow part is 30 square centimeters, and the area of the trapezoid is how many square centimeters?

Trapezoid bottom: 30 × 2 △ 6, = 60 △ 6, = 10 (CM), trapezoid area: (4 + 10) × 6 △ 2, = 14 × 6 △ 2, = 42 (square cm); answer: trapezoid area is 42 square cm

If the negative numbers a, B, C satisfy a + B + C = - 1, then the maximum value of 1 △ a + 1 △ B + 1 △ C

Can you tell me why three negative numbers add up to a positive number?

Using the invariance of differential form to find the differential of function y = cosln (x ^ 2 + e ^ - 1 / x)

Let ln (x ^ 2 + e ^ (- 1 / x)) = U
Let x ^ 2 + e ^ (- 1 / x) = t
y=cosu
dy=-sinudu
u=lnt
du=1/t*dt
dt=2x+e^(-1/x)/x^2*dx
So dy = - sinln (x ^ 2 + e ^ (- 1 / x)) * (2x + e ^ (- 1 / x) / x ^ 2) / (x ^ 2 + e ^ (- 1 / x)) DX

The sum of length and width of a rectangular vegetable field is 8 meters. How many meters is the perimeter of this rectangular vegetable field?

8 × 2 = 16 meters. A: the perimeter of this vegetable field is 16 meters

A's Cube - A's square B + AB's square + A's Square b-ab's square + B's cube = merge congeners if a = - 3, B = 2, the value is?

a³-a²b+a²b²+a²b-a²b²+b³
(remove everything in the middle and leave a and b)
The original solution = A & # 179; + B & # 179;
= -3³+2³
= -27+8
= -19

If f (x) is a continuous odd function, what is the definite integral ∫ f (x) DX in the integral interval [- 2,2]?

The definite integral of an odd function on a symmetric interval must be 0, which will be proved in books,