Example 1: The distance between the two places is 300 kilometers. When a ship sails between two places, what are the speed of the ship in still water and the speed of the current if it is downstream 15 hours and upstream for 20 hours?

Analysis: downstream speed = ship speed+water speed; Countercurrent speed+ship speed-water speed

Solution: This problem can be easily solved with the idea of equation.

Let the water speed be xKm/h and the ship speed be ykm/h.

Equations can be listed from the meaning of the question.

300/(x+y)= 15

300/(y-x)=20

Where x+y is the downstream speed (speed = water speed+ship speed)

X-y is the speed of countercurrent (speed = ship speed-water speed)

By solving a quadratic equation with one variable,

X = 2.5km/h was found.

Y =17.5km/h

Example 2: When a ship sails in A and B, the time required for sailing with the current is four fifths of that required for sailing with the current. A ship is sailing at a speed of 65,438+08 kilometers per hour in still water, so as to get the current speed.

Solution: If the current velocity is X, the downstream velocity is x+ 18, and the upstream velocity is18-X.

Assume that the upstream sailing time is t and the downstream sailing time is 4t/5.

There is (x+18) 4t/5 = (18-x) t.

Get x=2.

Example 3: (MX+8) (2-3x) expansion does not contain X term, so find the value of m.

Solution: ∫ (MX+8) (2-3x) = 2mx-3mx2+16-24x =-3mx2+(2m-24) x+16,

And the result does not contain x items,

∴2m-24=0，

∴m= 12。

Example 4

(x？ +mx+8)(x？ After -3x+n) expansion, find the quadratic term and cubic term without x, and find the value of m+n.

Solution: ∫(x? +mx+8)(x？ -3x+n)

= the 4th power of X-3x+the 3rd power of NX? +the third power of MX -3mx? +mnx+8x？ -24x+8n

= the 4th power of x-the 3rd power of (3-m) x +(n-3m+8)x? +(mn-24)x+8n

Because there are no quadratic terms and cubic terms,

So -(3-m)=0, and n-3m+8=0.

So m = 3 and n = 1.

m+n=4