Verification: any function whose domain of definition is r can be expressed as the sum of an odd function and an even function

Verification: any function whose domain of definition is r can be expressed as the sum of an odd function and an even function

It is proved that the function f (x) whose domain of definition is r can be expressed as the sum of an odd function g (x) and an even function H (x)
∴f(x)=g(x)+h(x).①
f(-x)=g(-x)+h(-x)
And G (- x) = - G (x), H (- x) = H (x)
∴f(-x)=-g(x)+h(x).②
It is known from (1) and (2) that h (x) = [f (x) + F (- x)] / 2, G (x) = [f (x) - f (- x)] / 2
Test: H (- x) = [f (- x) + F (x)] / 2 = H (x)
g(x)=[f(-x)-f(x)]/2=-g(-x)
The function f (x) with the domain R can be expressed as the sum of an odd function g (x) and an even function H (x)
And H (x) = [f (x) + F (- x)] / 2, G (x) = [f (x) - f (- x)] / 2

For the even function f (x) with the domain R and the odd function g (x) with the domain R, there are ()
A. f（-x）-f（x）＞0B. g（-x）-g（x）＞0C. g（-x）g（x）≥0D. f（-x）g（-x）+f（x）g（x）=0

∵ f (x) is an even function, G (x) is an odd function, Let f (x) = f (x) g (x), then f (- x) = f (- x) g (- x) = f (x) [- G (x)] = - f (x) g (x) ∵ f (- x) g (- x) + F (x) g (x) = 0, so D is selected

To process a batch of parts, it takes 20 days for Party A to do it alone and 30 days for Party B to do it alone. Party A and Party B work together for 6 days and process 1500 parts
How many pieces are there?

A's work efficiency: 1 / 20
Efficiency of Party B: 1 / 30
6 * [(1 / 20) + (1 / 30)} = 1 / 2
Total quantity of parts: 1500 / (1 / 2) = 3000 (pieces)

（9.8-1.8x）×7=18.22.6x=51.6-3.4x3（4x-2）-2（3x+3）=9-8x20+4x=6x-245（x-1）=x+13（3x-2）=10-0.5（x-3.5）0.4（x-0.2）+1.5=0.7x-0.38（0.6x+420）÷（x+20）=335（2-x）+15（6-5x）=22x+7+21（4-3x）

（1）（9.8-1.8x）×7=18.2，  （9.8-1.8x）×7÷7=18.2÷7，            9.8-1.8x=2.6，       9.8-1.8x+1.8x=2.6+1.8x，            2.6+1.8x=9.8，        2.6+1.8x-2.6=9.8-2.6，                1.8x=7.2，           1.8x÷1.8=7.2÷1.8，                   x=4；（2）2.6x=51.6-3.4x，2.6x+3.4x=51.6-3.4x+3.4x，       6x=51.6，    6x÷6=51.6÷6，        x=8.6；（3）3（4x-2）-2（3x+3）=9-8x，                  6x-12=9-8x，               6x-12+8x=9-8x+8x，                 14x-12=9，              14x-12+12=9+12，                     14x=21，                14x÷14=21÷14，                       x=1.5；（4）20+4x=6x-24，  20+4x-4x=6x-24-4x，     2x-24=20，  2x-24+24=20+24，        2x=44，     2x÷2=44÷2，         x=22；（5）5（x-1）=x+1，        5x-5=x+1，       5x-5-x=x+1-x，        4x-5=1，      4x-5+5=1+5，          4x=6，       4x÷4=6÷4，           x=1.5；（6）3（3x-2）=10-0.5（x-3.5），         9x-6=11.75-0.5x，    9x-6+0.5x=11.75-0.5x+0.5x，       9.5x-6=11.75，     9.5x-6+6=11.75+6，         9.5x=17.75，    9.5x÷9.5=17.75÷9.5，            x=4738；（7）0.4（x-0.2）+1.5=0.7x-0.38，           0.4x+1.42=0.7x-0.38，      0.4x+1.42-0.4x=0.7x-0.38-0.4x，            0.3x-0.38=1.42，      0.3x-0.38+0.38=1.42+0.38，                0.3x=1.8，           0.3x÷0.3=1.8÷0.3，                    x=6；（8）（0.6x+420）÷（x+20）=3，（0.6x+420）÷（x+20）×（x+20）=3×（x+20），                      3（x+20）=0.6x+420，                           3x+60=0.6x+420，                     3x+60-0.6x=0.6x+420-0.6x，                         2.4x+60=420，                     2.4x+60-60=420-60，                            2.4x=360，                      2.4x÷2.4=360÷2.4，                              x=150；（9）35（2-x）+15（6-5x）=22x+7+21（4-3x），                160-110x=91-41x，           160-110x+110x=91-41x+110x，                   69x+91=160，               69x+91-91=160-91，                     69x=69，                  69x÷69=69÷69，                       x=1．

When 1 / 10 of the grain stored in warehouse B is transported away, the grain stored in warehouse A and warehouse B is equal, and the weight ratio of the original grain stored in warehouse A and warehouse B is ()

Because 0.9 * b = a
So a / b = 0.9 / 1 = 9 / 10
That is, a: B = 9:10

It is necessary to solve the equation 1 / 3x = 2 / 5Y

Multiply both sides by 15 to get
5x=6y
y=5x/6

1998÷199819981999．

1998÷199819981999，=1998÷1998×1999+19981999，=1998÷1998×20001999，=1998×19991998×2000，=19992000．

It is known that the equal ratio sequence {an} has m terms (M is greater than or equal to 3), and each term is positive, A1 = 1, a1 + A2 + a3 = 7
It is known that the equal ratio sequence {an} has m terms (M is greater than or equal to 3), and each term is positive, A1 = 1, a1 + A2 + a3 = 7. Find the general term an? Of the sequence {an}?

A1 = 1, A2 = q, A3 = q ^ 2, then a1 + A2 + a3 = 1 + Q + Q ^ 2 = 7, namely Q ^ 2 + q-6 = 0, the solution is q = 2 or q = - 3 (rounding off), so q = 2, so an = A1 × Q ^ (n-1) = 2 ^ (n-1)

Simple operation of 97 out of 197 * 98 + 1 out of 98

197 * 97 / 98 + 1 / 98
=197*97/98+1-97/98
=97/98*(197-1)+1
=97/98*196+1
=97*2+1
=(100-3)*2+1
=200-6+1
=195

As shown in the figure, e and F are the midpoint of the edge AD and BC of rectangle ABCD respectively. If rectangle ABCD is similar to rectangle eabf, ab = 10, calculate the area of rectangle ABCD

Let BC length x, because rectangle ABCD and rectangle eabf are similar, then x / 10 = 10 / (0.5x), the solution is x = 10 √ 2, so rectangle ABCD area = 10x = 100 √ 2 = 141.42