Grade five and grade six of sunshine primary school have 208 students, twice less than 2

Grade five and grade six of sunshine primary school have 208 students, twice less than 2

Suppose the number of students in grade five is X
1.1X－2+X＝208
The solution is: x = 100
So the number of sixth grade students is 108
A: there are 100 students in grade five and 108 students in grade six

There are 102 students in class A and class B in Grade 6 of Guangming primary school. If one eighteenth of class A is transferred to class B, the number of students in the two classes is equal
How many students are there in class a?
Do me a favor

Equal number of people: 102 / 2 = 51 (people)
A: 51 / (1-1 / 18) = 54 (people)

The number of students in class A is 1.4 times that of class B. if nine students are transferred from Class A to class B, the number of students in the two classes will be equal. How many students are there in class A and class B?
Please write the formula clearly! I hope to answer the question quickly!

If there are x people in class B, then there are 1.4x people in class A, so x + 9 = 1.4x-9, 9 + 9 = 1.4x-x, 18 = 0.4x

Given x > 1, prove the inequality x > in (1 + x)

Let f (x) = x-ln (1 + x) f '(x) = 1-1 / (x + 1) = x / (x + 1)
Because f (1) = 1-ln2 = lne-ln2 > 0 (y = LNX is monotone increasing function in x > 0, E > 2) when x > 1, f '(x) > 0, the derivative of F (x) is positive, so f (x) is monotone increasing function in x > 1 interval, and f (1) > 0, so when x > 1, f (x) > 0
That is x-ln (1 + x) > 0 x > ln (1 + x)

1 - 1 / 1-x - x squared - 2x + 1 / X squared - 1 =?
1 - (1 / 1-x) - (square of x-2x + square of 1 / x-1)=

1-（1/x-1）-(x²-2x+1/x²-1)
=1-（1/x-1）-（(x+1)²/(x+1)(x-1))
=（x-1/x-1)-(x-1/1）-（x+1/x-1)
=-1/x-1

As shown in the figure, in known parallelogram ABCD, AE and CF are bisectors of ∠ DAB and ∠ BCD, respectively

A quadrilateral ABCD is a parallelogram
∴AB=CD,AD=BC,AB//CD
∴∠BAE=∠DEA
∵ AE bisection ∠ DAB
∴∠BAE=∠DAE
∴∠DAE=∠DEA
∴AD=DE
Similarly, BF = BC
∴DE=BF
∴AB-BF=CD-DE
That is AF = CE
∵AF//CE
The quadrilateral afce is a parallelogram

Given that positive real numbers a and B satisfy a + B = 1, find the maximum or minimum of a / (1 + b) + B / (1 + a)

Solution
It's all right
a/(1+b)+b/(1+a)=(a+a^2+b+b^2)/(1+a+b+ab)
Take a + B = 1 and a ^ 2 + B ^ 2 = (a + b) ^ 2-2ab = 1-2ab into the above formula
The above formula = (2-2ab) / (2 + AB)
=[-2*(2+ab)+6]/(2+ab)
=-2+6/(2+ab)
And 1 = a + b > = 2 √ (AB)
So AB = - 2 + 6 / (2 + 1 / 4) = 2 / 3
So the minimum value is 2 / 3 when a = b = 1 / 2
And ab > 0
So the above formula

A calculus problem. Y = SiNx, find dy / DX, hope to provide process

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In the parallelogram ABCD, AC = √ 65, BD = √ 17, the perimeter is 18, then the area of the parallelogram is ()

In the parallelogram ABCD, AC = √ 65, BD = √ 17, perimeter 18, then the area of this parallelogram is (16). Let AB = CD = a, ad = BC = B, from perimeter 18, then a + B = 9, so a & sup2; + B & sup2; + 2Ab = 81 has cosine theorem, and AC & sup2; + BD & sup2; = A & sup2; + B & sup2; + A & sup2; + B & sup2; AC =

Given the function f (x) = log3x + 2 & nbsp; (x ∈ [1,9]), then the maximum value of function y = [f (x)] 2 + F (x2) is ()
A. 13B. 16C. 18D. 22

The definition field of function y = [f (x)] 2 + F (x2) is {x | 1 ≤ x ≤ 9 and 1 ≤ x2 ≤ 9} = [1,3] and y = [f (x)] 2 + F (x2) = (log3x + 2) 2 + log3 (x2) + 2 = (log3x) 2 + 6log3x + 6. Let t = log3x, ∵ x ∈ [1,3], ∵ t ∈ [0,1] ∵ y = T2 + 6T + 6 = (T + 3) 2-3 be monotone on [0,1]