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f'(x)=e^x*1/x+e^x*lnx-e^x+1f'(1)=e+0-e+1=1>0

Let g(x)=[ f'(x)-1 ]/e^x=1/x+lnx-1g'(x)=1 x * 1-1 x), at [1,e] Evergrande is 0f'(x)-1 ] e x is monotonically increasing at [1,e], hence f'(x) is also monotonically incremental, f'(x) >0, there is no such x

Exist. The tangent is perpendicular to the y-axis, that is, the slope of the tangent is 0, which is f'(x)=0f'(x) = (x x) * (1 + lnx) -e x + 1 because f'(1)<0,f'(e)>0

So there is 1

f'(x)=e x*1 x+e x*lnx-e x+1 to prove that the tangent of f(x) is perpendicular to the y-axis, that is, to prove that the derivative of f(x) can be equal to 0, and to bring 1 and e respectively to the derivative, if the different signs show that there is a zero point between [1,e].

That is, it will be proven.

Solution: f(x)=e x*lnx-e x+x

Derivative f'(x)=e x*(lnx+1 x-1)+1, let h(x)=lnx+1 x-1

f'(x)=e^x*h(x)+1

Derivative h'(x)=1/x-1/x^2

When x belongs to [1,e], h'(x)>=0

So h(x) increases monotonically.

So f'(x) Monotonous increase.

Seek f'(1)=1

So when x belongs to [1,e], f'(x)>=1So when x belongs to [1,e], f(x) the slope of the tangent at each point "=1, and there is no tangent perpendicular to the y-axis.

Using the cosine theorem, cosa = (ac + ab -bc ) (2·ac·ab) = (1+3-4) (2·1·2).

0 so a = 2 (or a = 90°).

bc²=ac²+ab²

abc is a right triangle, a is a right angle, a = 90 °

If the speed of the fast train is x, the speed of the slow train is 5/7x.

Column equation: 4x-48 = 4 * 5/7 x + 48. This can be solved to x= 84.

Let AB and the two places be separated by S kilometers.

s=（4*84-48）*2=476

1 = or years old sin = -1sin alpha = -1, cos beta = -1, Ahu sheng line laughing hand erfa = beta = 3 2

cos (alpha + beta) = -1

In summary, cos (alpha + beta) = -1

Let them be x, y, z, respectively, and the mask can be solved as the stuffy group below the macrobi.

x + y + z =180°

x + z = 2y

Y = 60°

The sum of the inner angles of the triangle is 180°, while the three inner angles of the triangle abc form a series of equal differences, dividing 180 by 3 = 60°

4+ tan²xcosx=4cosx

4+sin²x/cos²x*cosx=4cosx4+sin²x/cosx=4cosx

4cosx+sin²x=4cos²x

4cosx+sin²x+cos²x=5cos²x4cosx+1=5cos²x

Let cosx=a

5a^2-4a-1=0

a1=-1/5,a2=1

Because of the range, 1 5

So cosx=-1 5

x = radians).

When x=-18 (, the function obtains a maximum value, and the maximum value is 142

Plug-in method.

Imagine 15 balls, 12 green balls and 3 red balls, lined up in a row, with the position of the balls representing their numerical size, where the red balls represent the set A

First, 0-5 green balls are inserted between A2 and A3, and when 5 green balls are inserted, there are 7 green balls left, and they form 8 gaps (including two at both ends), and A1, A2 to A3 are considered as two plates, and 8 voids are inserted, and their combined number is 28

When inserting 4 green balls, there are 9 gaps and the number of combinations is 36

3 when inserted 45

55 when inserted 2

66 when inserting 1

78 when 0 is inserted

In summary, there are 308 species in total.

1.Let f(x)=x + (a-3)x+1, then f(2)<0, so b2Use the root finding formula to find the two, the big one is less than 2, the small one is greater than 3, and the equal sign is verified, and it's OK.

3], and a b = [-1,2), so x + px + q = 0 has a stick of 2, so 2p + q = -4

4.When -1<=x<=1, meta inequality |(x+1)/(x-1)|=|1+ 2/x-1|<1, so -2<2 x-1<0, x<-1

A is sufficiently unnecessary for B.

It is known that f(x) = sinx

When x=x-2, f(x)=f(x- 2)=sin(x- 2)=sin(x-2 +3 2)=-cosx

It is faster to solve this problem by combining numbers and shapes.

First, draw a curve with f(x)=sinx (a sinusoidal curve with a maximum of =1 over the origin, t=2, y).

When the function changes from f(x) to f(x-2), the sinusoidal curve graph remains the same longitudinally and shifts horizontally to the right by 2 units, which happens to become an inverse cosine curve.

Question 2: I've saved all the degrees

cos(-820)=cos(-360*2-100)=cos(-100)=t

sin(-100) = (1-t) under the root number (1-t)).

then tan(-100)=[sin(-100)] [cos(-100)]= (1-t) t

tan(-100)=tan(-100+360+180)=tan(440)=½(1-t²)/t