Given that e is the base of natural logarithm, the function f (x) = ax ^ 2 / e ^ x, where a belongs to R and a is not equal to 0 When a is greater than 0, the maximum value of F (x) is 1 / E

F '(x) = ax (2-x) / e ^ x > 0, when a > 0, the interval is (0,2); when A0, f' (x) = 0 = > x = 0 or 2 X2, the function decreases, which can be judged by drawing, when x = 2, f (2) = 1 / E, a

Let f (x) be continuous on [a, b], and (a, b) be differentiable, and f (a) = 0. It is proved that there is at least one point ξ∈ (a)
Let f (x) be continuous on [a, b], and (a, b) be differentiable, and f (a) = 0. It is proved that there is at least one point ξ ∈ (a, b), such that f (ξ) = (B - ξ) * f '(ξ)

Let f (x) = (X-B) * f (x) be known that f (x) is differentiable and continuous in the interval [A & nbsp; b], and then f (x) is suitable for Rolle's theorem, that is, there is a point ξ, such that F & # 39; (ξ) = 0f & # 39; (x) = f (x) + (X-B) F & nbsp; &# 39; (x) F & # 39; (ξ) = F & # 39; (x) = (X-B) F & nbsp; &# 39; (x); (ξ) + (ξ - b) F & nbsp; ' (ξ) = 0 is reduced to (ξ) = (B - ξ) F & #39; (ξ) and in practice * generally need not be written,

The curve t equation is x2 / 4 + y2 = 1
Let the line L passing (0, - 2) intersect with the curve t at two points c d, and the vector OC multiplied by the vector od equals 0 (o is the origin of the coordinate)?
Let the equation of line l be kx-y-2 = 0 and the curve T, then (1 + 4k2) x2-16kx + 12 = 0
Then I use Veda's theorem to find X1 + X2, x1x2, and then I don't know how to go on
I know these questions are to be replaced, but I just don't know how to replace them
cry for help
X2 is the square of X, and the others are the same

Because the vector OC times the vector od equals 0, OC is perpendicular to OD
That is, (Y1 / x1) (Y2 / x2) = - 1, Y1 and Y2 can be replaced by X1 and x2

What are the four operations?

Addition, subtraction, multiplication and division

lg2x+（lg2+lg3）lgx+lg2•lg3=0 x1*x2=?
If you change the title to lg2x + (LG2 + lg5) lgx + LG2 &; Lg3 = 0 & nbsp; & nbsp; and you can't use cross multiplication, what is the general solution?

1、(lgx+lg2)(lgx+lg3)=0
∴lgx=-lg2 lgx=-lg3
∴x=1/2 x=1/3
If you change it to lg5, you need to use the root formula to find the answer. The answer is very boring

Draw the largest square in the same circle. The diagonal of the square is 8 cm. The area of the circle is () square cm more than that of the square

The radius of a circle is four centimeters, and the area of a square is 32 square centimeters

Read, think and write.
（1）zhang lan is going to hainan.lt is hot and sunny.she is going shopping.what is she going to buy?
_____________________
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（2）tom is going to harbin .lt is cold and snowy.he is going shopping.what is he going to buy?
_
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_____________________

Child paper. We know where you are god horse topic, at least type up the topic to have a look

If the algebraic formula 2x ^ 2-3x + k-kx ^ 2 + 4kx-4 is a quadratic binomial with no constant term and only two terms
Polynomial, then the polynomial can be simplified as?

2X ^ 2-3x + k-kx ^ 2 + 4kx-4 is a quadratic binomial with no constant term and only two terms
So K-4 = 0
k=4
The original formula can be changed to: 2x ^ 2-3x-4x ^ 2 + 16x = - 2x ^ 2 + 13X

If the equation sin π x2 = logax (a > 0 and a ≠ 1) has exactly three unequal real roots, then ()
A. B. a ∈ (5,9) C. A ∈ (17,13) d. a ∈ (17,13) ∪ (5,9)

Let f (x) = sin (π 2x) and G (x) = logax. The equation sin π x2 = logax (a ＞ 0 and a ≠ 1) has exactly three unequal real roots. It can be understood that the graph G (x) of function f (x) has exactly three common points. The graph of the two functions is as follows: when a ＞ 1, in the case of a = 5, the logarithmic function y = g (...)

Simple calculation questions for grade five
Ten questions

102×4.5
7.8×6.9＋2.2×6.9
5.6×0.25
8×（20－1.25）
1）127+352+73+44 （2）89+276+135+33
（1）25+71+75+29 +88 （2）243+89+111+57
9405-2940÷28×21
920-1680÷40÷7
690+47×52-398
148+3328÷64-75
360×24÷32+730
2100-94+48×54
51+（2304-2042）×23
4215+（4361-716）÷81
（247+18）×27÷25
36-720÷（360÷18）
1080÷（63-54）×80
（528+912）×5-6178
8528÷41×38-904
264+318-8280÷69
（174+209）×26- 9000
814-（278+322）÷15
1406+735×9÷45
3168-7828÷38+504
796-5040÷（630÷7）
285+（3000-372）÷36
1+5/6-19/12
3x(-9)+7x(-9
(-54)x1/6x(-1/3)
1.18.1＋（3－0.299÷0.23）×1
2.(6.8-6.8×0.55)÷8.5
3.0.12× 4.8÷0.12×4.8
4.（3.2×1.5+2.5）÷1.6 （2）3.2×（1.5+2.5）÷1.6
5.6-1.6÷4= 5.38+7.85-5.37=
6.7.2÷0.8-1.2×5= 6-1.19×3-0.43=
7.6.5×（4.8-1.2×4）= 0.68×1.9+0.32×1.9
8.10.15-10.75×0.4-5.7
9.5.8×（3.87-0.13）+4.2×3.74
10.32.52-（6+9.728÷3.2）×2.5
11.[（7.1-5.6）×0.9-1.15] ÷2.5
12.5.4÷[2.6×（3.7-2.9）+0.62]
13.12×6÷（12-7.2）-6
14.12×6÷7.2-6
15.33.02－（148.4－90.85）÷2.5
7×(5/21+9/714)

Given that e is the base of natural logarithm, the function f (x) = ax ^ 2 / e ^ x, where a belongs to R and a is not equal to 0 When a is greater than 0, the maximum value of F (x) is 1 / E