6.7x-60.3=6.7 x=

6.7x-60.3=6.7 x=

6.7x-60.3=6.7
6.7x=6.7+60.3
6.7x=67
x=67÷6.7
x=10

How to write 0.6 / 5x-1 = 1-0.3 / 4-7x? Fast

5x-1 in 0.6 = 4-7x in 1-0.3
5x-0.6 = 0.6-8-4.2x (multiply both sides of the equation by 0.6)
5X+4.2X=0.6-8+0.6
9.2X= -6.8
X= -17/23
X = 17 out of 23

The speed of car a is 78 times of that of car B. the two cars are facing each other at the same time from the two places of AB, and they meet at the distance of 4km from the midpoint. The distance between the two places is______ Kilometers

(4 × 2) △ 88 + 7-78 + 7, = 8 △ 815-715, = 8 △ 115, = 8 × 15, = 120 (km); answer: the distance between the two places is 120 km. So the answer is: 120

It is proved that when n > 2, 1 + 1 / 2 + 1 / 3 +... + 1 / 2 ^ n (7n + 11) / 12

It can be proved by mathematical induction
When n = 2, left = 1 + 1 / 2 + 1 / 3 + 1 / 4 = 25 / 12, right = (7 × 2 + 11) 12 = 25 / 12. The original inequality holds
If n = K (K ≥ 2), the proposition holds, that is, 1 + 1 / 2 + 1 / 3 + +1/2^k≥(7k+11)/12
So, when n = K + 1
1+1/2+1/3+… +1/2^(k+1)≥(7k+11)/12+1/(2^k+1)+1/(2^k+2)+… +1/2^(k+1)
And 1 / (2 ^ k + 1) + 1 / (2 ^ k + 2) + +1/2^(k+1)
=1/(2^k+1)+1/(2^k+2)+… +1/[2^k+2^(k-1)]+1/[(2^k+2^(k-1)+1]+1/[(2^k+2^(k-1)+2]+… +1/2^(k+1)
≥2^(k-1)/[2^k+2^(k-1)]+2^(k-1)/2^(k+1)=7/12
So, 1 + 1 / 2 + 1 / 3 + +1/2^(k+1)≥(7k+11)/12+7/12=[7(k+1)+11]/12
Therefore, the proposition holds when n = K + 1
So for all n ≥ 2, the inequality 1 + 1 / 2 + 1 / 3 +... + 1 / 2 ^ n ≥ (7n + 11) / 12 holds

On a 1:6000000 map, the distance between Nanjing and Beijing is 15cm,
If we draw the distance between Nanjing and Beijing on a map with a scale of 1:5000000, how many centimeters should we draw? Find the equation

The actual distance is: 15 △ 1 / 6000000 = 90000000 cm
If we draw the distance between Nanjing and Beijing on a map with a scale of 1:5000000, we should draw:
90000000 × 1 / 5000000 = 18cm
If you don't understand, please ask

Given f (x) = x ^ 2 + PX + Q, it is proved that at least one of {f (1)}, {f (2)}, {f (3)} is not less than 1 / 2
Prove it by disprovement
I hope there is a specific process and explanation,
Note that "{}" stands for "absolute value"

This should be counter evidence
f(1)=p+q+1,f(2)=2p+q+4,f(3)=3p+q+9.
Suppose that | f (1) |, | f (2) |, | f (3) | are less than 1 / 2, then | f (1) - 2F (2) + F (3) | ≤ | f (1) | + 2 | + | f (3)|

2. When a fleet transports a pile of coal, it transports 1 / 6 of the pile of coal on the first day and 30 tons more on the second day than on the first day. At this time, the ratio of the transported coal to the remaining coal tons is 7:5. How many tons of the pile of coal are there in total?

Suppose there are x tons of coal in this pile
1/3X+30:2/3X=7:5
The solution is x = 50
There are 50 tons of coal in this pile

It is known that the general formula of the arithmetic sequence {an} is an = 2N-1, and the general formula of the arithmetic sequence {BN} is BN = 3 ^ (n-1). Let a sequence have C1 / B1 + C2 / B2 + C3 / B3 + for any natural number n +Cn / BN = a (n + 1) holds, C1 + C2 + C3 + +The value of c2007

The speed ratio of car a and car B is 5:8. The two cars start from ab at the same time, and then meet 24 kilometers away from the midpoint. How far is the distance between the two places?
It must be solved in proportion

The speed ratio is the distance ratio
This is the key to solving the problem
So the distance between the two places is a kilometer
（1/2a-24）：（1/2a+24）=5：8
4a-192=2.5a+120
1.5a=312
A = 208 km
The distance between the two places is 208 km

If the solution set of the inequality ax + 2 greater than 0 is (negative infinity, 1), then the real number a is?

ax+2>0
ax>-2
1)a>0,x>-2/a
2) A = 0, X belongs to R
3)x