If the tangent of the square of the curve y = ax - ax + 1 at the point (1,0) is perpendicular to the straight line 2x + y + 1 = 0, then a is equal to? Wrong, at (0,1)

If the tangent of the square of the curve y = ax - ax + 1 at the point (1,0) is perpendicular to the straight line 2x + y + 1 = 0, then a is equal to? Wrong, at (0,1)

y=ax^2-ax+1
Y '= 2ax-a, x = 1, y' = 2a-a = a tangent: y = a (x-1)
2x+y+1=0 y=-2x-1
-2*a=-1
a=1/2

Let the tangent of the curve y = AX2 at point (1, a) be parallel to the straight line 2x-y-6 = 0, then the value of a is______ ．

Y '= 2aX, then the slope of the tangent k = y' | x = 1 = 2A, ∵ the tangent is parallel to the straight line 2x-y-6 = 0 ∵ 2A = 2 ∵ a = 1, so the answer is 1

If the tangent of the curve y = x + 1x − 1 at point (3, 2) is perpendicular to the straight line ax + y + 1 = 0, then the value of a is______ ．

The derivative of function y = x + 1x − 1 = 1 + 2x − 1 & nbsp; is y ′ = − 2 (x − 1) 2, the tangent slope of curve y = x + 1x − 1 at point (3, 2) is - 12, from - 12 × (- a) = - 1, a = - 2, so the answer is: - 2

lim(x→+∞)(x+e^x)^(1/x)

The answers are as follows:

Calculate 1-2 + 3-4 + 5-6 +... + 199-200 =?

1-2+3-4+5-6+...+199-200
=（1-2）+（3-4）+（5-6）+...+（199-200）
=（-1）+（-1）+（-1）+.+（-1）
=（-1）x100
= -100

LIM (1 / sinx-1 / x) (x tends to 0)
When we find the solution, we first divide it. The molecule is x-sinx. Can we use the equivalent infinitesimal to change it to x-x
Why is that not right

1/sinx-1/x=（x-sinx）/xsinx
Seeking derivative of numerator denominator
（1-cosx）/（sinx+xcosx）
Seeking derivative again
sinx/（cosx+cosx-xsinx）
X tends to zero
The above formula = 0 / (1 + 1-0) = 0
LIM (1 / sinx-1 / x) (x tends to 0) = 0

How to calculate the following question: 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + 4 × 5 × 6 + 5 × 6 × 7 + 6 × 7 × 8 + 7 × 8 × 9 + +n×(n+1)×(n+2)

(n-1) (n + 1) = n ^ 2-1, then 1 * 2 * 3 = 2 ^ 3-2 * 3 * 4 = 3 ^ 3-3. Then the original formula = 2 ^ 3 + 3 ^ 3 + 4 ^ 3 +... + (n + 1) ^ 3-2-3-4 -... (n + 1) 1 ^ 3 + 2 ^ 3 +... + n ^ 3 = [n (n + 1) / 2] ^ 2, so 2 ^ 3 + 3 ^ 3 + 4 ^ 3 +... + (n + 1) ^ 3 = [(n + 1) (n + 2) / 2] ^ 2-1 2 + 3 + 4 +... + + (n + 1) = (2 + N + 1) n / 2, so the original formula is

(ex * sinx-x (1 + x)) / (x2 * SiNx) how to find the limit at 0 with Taylor formula?

Expand all functions at 0
Then add and multiply, and keep about 3 items
It becomes the limit of division of two polynomials

Assembly language programming calculation s = 1 + 2 * 3 + 3 * 4 + 4 * 5 +. + n * (n + 1) until n > 200

N> That is to say, the last one is 201 * 202, which is the sum of 201 items
mov bx,1
mov dx,0
mov ah,2
mov cx,200
s:
mov al,ah
inc ah
mov si,ax
mul ah
add bx,ax
adc dx,0
mov ax,si
loop s

The limit x tends to 0 (E ＾ x-e ＾ - x) / SiNx = can we do it without Taylor formula

lim(x->0) (e^x-e^(-x))/sinx
=LIM (x - > 0) (e ^ x + e ^ (- x)) / cosx
=2