If a (2,2) B (a, 0) C (0, b), a > 0, b > 0 are collinear, find the minimum value of ab

If a (2,2) B (a, 0) C (0, b), a > 0, b > 0 are collinear, find the minimum value of ab

AB = (a-1,1), AC = (- b-1,2), from a, B, C collinear so AB, AC collinear so: (A-1) * 2 = 1 * (- B-1) get: B = 1-2a, ab = a-2a ^ 2. Because a > 0, so when a = 1 / 4, AB maximum is 1 / 8. At this time, B = 1 / 2 meets the condition
Typing is not easy,

Minmin lost the decimal point when reading a decimal, and the result was 204008. The original decimal was read only as zero, and the original decimal was zero______ Or______ ．

If you want to read only one zero on the decimal point, you can put the decimal point in front of 8 or 4, that is 2040.8 or 20.408

A and B start from ab at the same time and run in opposite directions. They meet at the 60th kilometer away from A. after meeting, they continue to move forward and reach each other's starting place

Assuming that the distance between a and B is h km, the speed of car a is x km / h, and the speed of car B is y km / h, the relationship is 1, (H + H-40) / x = (H + 40) / Y2, 60 / x = (H-60) y, and the calculation result is 110 km

Rational number addition and subtraction in junior high school
-{+ [- (+ 5)]} and
|1/3-1/2|+|1/4-1/3|+|1/5-1/4|+···+|1/10-1/9|.

Question 1: if there are two negatives and two negatives, the answer is 5
Question 2: because the number in each absolute value is negative, we should reverse the order of absolute value
Original formula = 1 / 2 [- 1 / 3 + 1 / 3-1 / 4 + ··· + 1 / 9] - 1 / 10
Offset before and after
=1/2-1/10
=2/5

The two trains leave 500 kilometers apart at the same time, but they don't meet each other four hours later. They are 20 kilometers apart. It is known that car a travels 65 kilometers per hour, and how many kilometers does car B travel per hour?

(500-20) △ 4 = 480 △ 4 = 120 (km); 120-65 = 55 (km); a: car B runs 55 km per hour

If the average number of a group of data 6,7,6,4, x, 3 is 5, then the number of this group of data is ()

6,7,6,4,x,3
There are 6 numbers in total
The average is five
So 6 + 7 + 6 + 4 + X + 3 = 5 × 6
x=4
The number of this set of data is 6, 7, 6, 4, 4, 3

The highway from city a to city B is 50 kilometers long. A rides a bicycle and B rides a motorcycle. B starts one and a half hours late and arrives one hour earlier. If the speed of the motorcycle is high, the speed of the motorcycle will be high
The road from city a to city B is 50 kilometers long. A rides a bicycle and B rides a motorcycle. B starts an hour and a half late and arrives an hour earlier. If the speed of the motorcycle is two and a half times that of the bicycle, how many kilometers are the speeds of a and B in an hour?

If the speed of bicycle is set at x km, the speed of motorcycle is 5 / 2x. The motorcycle starts 1.5 hours later and arrives 1 hour earlier, so the time of bicycle is 1.5 + 1 = 2.5 hours,
50/x-50/(5x/2)=3/2+1,
X = 12, a bike 12 km / h;
12 * 5 / 2 = 30, B motorcycle speed 30 km / h

The sum of two natural numbers is 1000 less than the product, and one of them is a complete square number

Let the two numbers be a and B
Then AB = a + B + 1000, that is (A-1) (B-1) = 1001 = 7 * 11 * 13
Let a and B be complete squares, let a = n ^ 2,
Then A-1 = n ^ 2-1 = (n + 1) (n-1)
n=12
Then a = 144, B = 8

The distance between the two places is 96 kilometers. A and B cars leave from the two places and meet in 4 / 5 hours. A car travels 54 kilometers per hour. How many kilometers does a car travel per hour? There must be a formula and process

96÷4/5-54
=120-54
=66 (km)

It is known that in ladder ABCD, ab ‖ DC, points E and F are the midpoint of AD and BC respectively. It is proved that: (1) EF ‖ ab ‖ DC; (2) EF = 12 (AB + DC)

Connect AF and extend BC at point g. ∵ ad ≌ BC ≌ DAF = g. in △ ADF and △ GCF, ∵ DAF = g ≌ DFA = cfgdf = FC ≌ ADF ≌ GCF, ≌ AF = FG, ad = CG. In addition,