Write two products and sum of 2008x & # 178; - 2007x-2009 = 0, x1 × x2_____ ,x1+x2_______ Let two of the equations 15x & # 178; + 20x-9 = 0 be X1, X2, then X1 & # 178; + x2 & # 178=_______

Write two products and sum of 2008x & # 178; - 2007x-2009 = 0, x1 × x2_____ ,x1+x2_______ Let two of the equations 15x & # 178; + 20x-9 = 0 be X1, X2, then X1 & # 178; + x2 & # 178=_______

Write two products and sum of 2008x & # 178; - 2007x-2009 = 0, x1 × x2_ =-2009/2008____ ,x1+x2_ =2007/2008______ Let two of the equations 15x & # 178; + 20x-9 = 0 be x1, X2, then X1 & # 178; + x2 & # 178; = (x1 + x2) & # 178; - 2x1x2 = (- 20 / 15) & # 178; - 2 × (- 9) / 15 = 16 / 9 + 6 / 5 = 1

Given that the equation AX ^ 2 + BX + C (a ≠ 0) has real roots X1 and X2, let P = X1 ^ 2010 + x2 ^ 2010, q = X1 ^ 2009 + x2 ^ 2009, r = X1 ^ 2008 + x2 ^ 2008
Then AP + BQ + CR =? Be more specific

It's zero
ap+bq+cr
= x1^2008 * (a * x1^2 + b * x1 + c) + x2^2008 * (a * x2^2 + b * x2 + c)
X1 and X2 are two roots, so the result in brackets is 0, and so are 0

The maximum acceleration of the fighter on the aircraft carrier is 5m / S2, and it takes 50M / s to take off. At present, the deck length of the aircraft carrier is 160m
Can the plane take off safely
The booster is installed to make the aircraft reach a certain speed immediately in t = 1.5 S. the mass of the aircraft is known to be 5.4 T. The average thrust of the aircraft subjected to the booster is calculated by ignoring the thrust during the main thrust period

No, because at = 50, a = 5, so t = 10, because s = 1 / 2, a * t * t = 250m, and the deck is only 160m, so it can't take off. The average thrust is not easy to calculate, because it involves the time acting on the aircraft. Is it the start, middle, or tail? The results are different

The equation that 1 / 2 minus 2 x equals 1 / 4

The paper bridge is made of one to three old newspapers. A small amount of transparent tape can span a 35 cm "Canyon" with a width of more than 10 cm
It can carry more than 250 grams

First, stick the newspaper with adhesive tape, then the bearing capacity will be greater

X + 5 out of 8 equals
fast

x ＋5x/8＝13x/8

The distance between the two men is 40 km. B starts for 1.5 hours before B starts. A is behind and B is in front. They walk in the same direction. A's speed is 8 km per hour,
B's speed is 6 kilometers per hour. How many hours can a catch up with B?

B starts first: (1.5 * 6 + 40) / (8-6) = 65
A starts first: (40-1.5 * 8) / (8-6) = 14

It is known that a 2-AB = 26, ab-b 2 = - 18. Find the values of a 2-B 2 and a 2-2ab + B 2

a2-b2=a2-ab+ab-b2=26+（-18）=8．a2-2ab+b2=a2-ab-（ab-b2）=26-（-18）=44．

Calculation of 39 2 / 3 × 40 1 / 3 by factorization
There is another question in the title: the 2000 power of 3 + the 1999 power of 6 × 3 minus the 2001 power of 3. I can only type it in Chinese,

39 2 / 3 × 40 1 / 3 = (40-1 / 3) (40 + 1 / 3) = 40 - (1 / 3) = 1599 + 8 / 9 3 ^ 2000 + 6 × 3 ^ 1999-3 ^ 2001 = 3 ^ 2000 + 2 × 3 ^ 2000-3 ^ 2001 = 3 × 3 ^ 2000-3 ^ 2001 = 3 ^ 2001-3 ^ 2001 = 0

The store shipped 100 kg of fruit, including 67 kg of apples. What percentage of the total weight of apples?

Apple = 67 △ 100 = 67%