The numbers A, B and C are 1 once. If the sum of twice the reciprocal of c and b is equal to three times the reciprocal of a, find a, b and C.

The first line: let A =x, B =(x+ 1) and C =(x+2).

2/(x+2)+ 1/(x+ 1)= 3/x

2x？ +x+x？ +2x=x？ +3x+2

x？ = 1

X= 1 or-1

The reciprocal of ∫B = 1/(x+ 1)

∴x≠- 1

∴x= 1

The unit number of two digits is 7. If you use ten digits to reverse the unit number, the ratio of the obtained two digits to the original two digits is 8: 3, and the original two digits are found.

Second ways

Let the number in the original two digits be x.

( 10X+7)/(70+X)=3/8

3(70+X)=8( 10X+7)

2 10+3X=80X+56

77X= 154

X=2

So the original two-digit number is 27.

A ship sailed from Port A to Port B, and when sailing on this route, it walked 3/5 of the whole journey at normal speed. After that, slow down 10 knots and keep driving at this speed to port B. In this way, the time of this voyage after deceleration is the same as that without deceleration. What is the normal speed of this ship on this route?

The normal speed of the third ship on this route is X.

3/5÷x =( 1-3/5)÷(x- 10)

3(x- 10)=2x

x=30

The normal speed of this ship on this route is 30 knots.

The distance between Party A and Party B is 125km. From Party A to Party B, some people ride cars and others ride bicycles. Bikes leave 4 hours earlier than cars and arrive 1/2 hours later. It is known that the ratio of cycling speed to cycling speed is 2: 5. What are the speeds of bicycles and cars?

Let the self-help speed be x km/h, then the riding speed is 5x/2km/h.

Then the bus ride time is:125 ÷ 5x/2 = 50/x.

Then there is the equation: 125/x-50/x=4.5 (according to the time difference between riding and riding).

The solution is x = 50/3 km/h.

Then the speed is: 5/2 * 50/3 =125/3 km/h.

A motorcade plans to transport m tons of goods in t days. If n tons have been transported, (n is less than m), the number of days of remaining goods transportation is t 1=__, and the average tonnage of goods shipped every day is a=____.

Daily freight volume: metric tons

Then the number of days required to transport the surplus goods is: (m-n) ÷ m/t = (m-n) * t/m.

A = metric tons

It takes the same time for a ship to sail 80 kilometers downstream and 60 kilometers upstream. Given the speed of the current of 3 km/h, find the speed of the ship in still water.

Let the speed of the ship in still water be x,

Then the downstream speed is x+3.

The upstream speed is x-3.

Then it is: 80/(x+3)=60/(x-3)

Solve the equation, x = 21km/h.

I bought a batch of T-shirts in March, with a total purchase price of 6.5438+0.2 million yuan. Because of the best selling, the store bought another batch of T-shirts of the same model in April, with a total purchase price of 187500 yuan, 0.5 times that of March. However, the purchase price increased by 5 yuan. At first, these two batches of T-shirts were sold at 180 yuan. By the beginning of May, the store sold the remaining 100 pieces at a 20% discount.

Suppose the purchase price of each piece in March is X yuan, then the purchase price of each piece in April is X+5 yuan.

So (12 *10000/x) * (3/2) * (x+5) =18.75 *10000.

Get X= 120 yuan.

And the total number of clothes is (12 *10000/x) * 5/2 = 2500 pieces.

Total income = 2400 *180+100 *180 * 80% = 446400 yuan.

So gross profit = 446400-120000-187500 =138900 yuan.

/2x=2/x+3

x/x+ 1=2x/3x+3 + 1

2/x- 1=4/x^2- 1

5/x^2+x - 1/x^-x=0

1/2x=2/x+3

Diagonal multiplication

4x=x+3

3x=3

x= 1

The fractional equation should be tested.

X= 1 is the solution of the equation.

x/(x+ 1)= 2x/(3x+3)+ 1

Multiply both sides by 3(x+ 1)

3x=2x+(3x+3)

3x=5x+3

2x=-3

x=-3/2

The fractional equation should be tested.

X=-3/2 is the solution of the equation.

2/x- 1=4/x^2- 1

Multiply both sides by (x+ 1)(x- 1)

2(x+ 1)=4

2x+2=4

2x=2

x= 1

The fractional equation should be tested.

After testing, x= 1 makes the denominator 0, which is the increase of roots and is discarded.

So the original equation has no solution.

5/x^2+x - 1/x^2-x=0

Multiply both sides by x(x+ 1)(x- 1).

5(x- 1)-(x+ 1)=0

5x-5-x- 1=0

4x=6

x=3/2

The fractional equation should be tested.

X=3/2 is the solution of the equation.

1/2x=2/x+3

Diagonal multiplication

4x=x+3

3x=3

x= 1

The fractional equation should be tested.

X= 1 is the solution of the equation.

x/(x+ 1)= 2x/(3x+3)+ 1

Multiply both sides by 3(x+ 1)

3x=2x+(3x+3)

3x=5x+3

2x=-3

x=-3/2

The fractional equation should be tested.

X=-3/2 is the solution of the equation.

2/x- 1=4/x^2- 1

Multiply both sides by (x+ 1)(x- 1)

2(x+ 1)=4

2x+2=4

2x=2

x= 1

The fractional equation should be tested.

After testing, x= 1 makes the denominator 0, which is the increase of roots and is discarded.

So the original equation has no solution.

5/x^2+x - 1/x^2-x=0

Multiply both sides by x(x+ 1)(x- 1).

5(x- 1)-(x+ 1)=0

5x-5-x- 1=0

4x=6

x=3/2

The fractional equation should be tested.

X=3/2 is the solution of the equation.

5x/(3x-4)= 1/(4-3x)-2

Multiply by 3x-4

5x =- 1-2(3x-4)=- 1-6x+8

1 1x=7

x=7/ 1 1

The fractional equation should be tested.

test

X=7/ 1 1 is the solution of the equation.

1/(x+2)+ 1/(x+7)= 1/(x+3)+ 1/(x+6)

Turn fractions into common denominators

(x+7+x+2)/(x+2)(x+7)=(x+6+x+3)/(x+3)(x+6)

(2x+9)/(x^2-9x+ 14)-(2x+9)/(x^2+9x+ 18)=0

(2x+9)[ 1/(x^2-9x+ 14)- 1/(x^2+9x+ 18)]=0

Because x 2-9x+ 14 is not equal to x 2+9x+ 18.

So1/(x2-9x+14)-1/(x2+9x+18) is not equal to 0.

So 2x+9=0.

x=-9/2

The fractional equation should be tested.

test

X=-9/2 is the solution of the equation.

7/(x^2+x)+ 1/(x^2-x)=6/(x^2- 1)

Multiply both sides by x(x+ 1)(x- 1).

7(x- 1)+(x+ 1)=6x

8x-6=6x

2x=6

x=3

The fractional equation should be tested.

X=3 is the solution of the equation.

Simplify the assessment. [x-1-(8/x+1)]/[x+3/x+1] where X=3- radical number 2.

[X- 1-(8/X+ 1)]/[(X+3)/(X+ 1)]

= {[(X- 1)(X+ 1)-8]/(X+ 1)}/[(X+3)/(X+ 1)]

=(X^2-9)/(X+3)

=(X+3)(X-3)/(X+3)

=X-3

=-root number 2

8/(4x^2- 1)+(2x+3)/( 1-2x)= 1

8/(4x^2- 1)-(2x+3)/(2x- 1)= 1

8/(4x^2- 1)-(2x+3)(2x+ 1)/(2x- 1)(2x+ 1)= 1

[8-(2x+3)(2x+ 1)]/(4x^2- 1)= 1

8-(4x^2+8x+3)=(4x^2- 1)

8x^2+8x-6=0

4x^2+4x-3=0

(2x+3)(2x- 1)=0

x 1=-3/2

x2= 1/2

Substitution test, x= 1/2 makes denominator 1-2x and 4x 2- 1 = 0. refuse

So the original equation solution: x=-3/2.

(x+ 1)/(x+2)+(x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6)

1- 1/(x+2)+ 1- 1/(x+7)= 1- 1/(x+3)+ 1- 1/(x+6)

- 1/(x+2)- 1/(x+7)=- 1/(x+3)- 1/(x+6)

1/(x+2)+ 1/(x+7)= 1/(x+3)+ 1/(x+6)

1/(x+2)- 1/(x+3)= 1/(x+6)- 1/(x+7)

(x+3-(x+2))/(x+2)(x+3)=(x+7-(x+6))/(x+6)(x+7)

1/(x+2)(x+3)= 1/(x+6)(x+7)

(x+2)(x+3)=(x+6)(x+7)

x^2+5x+6=x^2+ 13x+42

8x=-36

x=-9/2

X=-9/2 is the root of the equation.

(2-x)/(x-3)+ 1/(3-x)= 1

(2-x)/(x-3)- 1/(x-3)= 1

(2-x- 1)/(x-3)= 1

1-x=x-3

x=2

The fractional equation should be tested.

X=2 is the root of the equation.