In △ ABC, note that a, B and C are opposite sides of a, B and C respectively, and S is the area of triangle, and prove that C ^ 2-A ^ 2-B ^ 2 + 4AB ≥ 4 √ 3S This is the 19th question of the test paper. It's not very difficult, but I haven't thought about it for a long time. Please help me solve it~ PS: because the characters are not standard, I'll use words to explain it again. Verification: the square of C minus the square of a minus the square of B plus 4AB is greater than or equal to 4 times the root sign 3 times s~

In △ ABC, note that a, B and C are opposite sides of a, B and C respectively, and S is the area of triangle, and prove that C ^ 2-A ^ 2-B ^ 2 + 4AB ≥ 4 √ 3S This is the 19th question of the test paper. It's not very difficult, but I haven't thought about it for a long time. Please help me solve it~ PS: because the characters are not standard, I'll use words to explain it again. Verification: the square of C minus the square of a minus the square of B plus 4AB is greater than or equal to 4 times the root sign 3 times s~

The cosine theorem C ^ 2 = a ^ 2 + B ^ 2-2abcosc is transformed into C ^ 2-A ^ 2-B ^ 2 + 4AB = 4ab-2abcosc triangle area s = 1 / 2absinc. Substituting into the inequality, 4ab-2abcosc ≥ 4 √ 3 × 1 / 2absinc is equivalent to 2-cosc ≥ √ 3sinc, i.e. 1 ≥ √ 3 / 2sinc + 1 / 2cosc, i.e. 1 ≥ sin (c + 30) is equivalent from top to bottom

a. B belongs to (0, positive infinity) 2C > A + B to prove C ^ 2 > ab

a>0,b>0
Then a + B ≥ 2 √ (AB)
2c>a+b
So 2C > 2 (√ AB)
4c^2>4ab
c^2>ab

Given a > 0, b > 0, 2C > A + B, prove the square of C > ab

Because 2C > A + B, a > 0, b > 0, a + b > = a * B under 2 radical, the proof of AB simultaneous square problem under 2C > 2 radical

Master Wang has to process 4080 parts, which is planned to be completed in 20 hours. After working as planned for 5 hours, his work efficiency has increased by 25%, which is ahead of schedule
After finishing the task, how many parts did Master Wang process per hour from the beginning to the completion of all tasks?

4080 / 20 = 204
204x (1 + 25%) = 255
(4080-204x5) / 255 = 12 hours
4080/（12+5）=240
An average of 240 parts are processed per hour

Master Zhang plans to process 1200 parts. As the work efficiency has been increased by 20%, it is completed one hour ahead of schedule. How many parts does Master Zhang plan to process per hour

xy=1200
x(1+20%)(y-1)=1200
The solution is x = 200, y = 6

Master Wang plans to process a batch of parts in two hours. When there are 160 parts left, the machine breaks down and the efficiency is 1 / 5 lower than the original plan. As a result, the task is delayed by 20 minutes. How many parts are there?

When 20 points = 13, 1-15 = 45160 ^ [13 ^ (1 ^ 45-1)] × 2, = 160 ^ [13 ^ (54 − 1)] × 2, = 160 ^ [13 ^ [14] × 2, = 160 ^ 43 × 2, = 120 × 2, = 240 (pieces); answer: there are 240 pieces of this batch of parts

Master Liu plans to process 640 parts in 4 days, and the actual work efficiency has been increased by 10%. How many parts are actually processed every day?

640÷4×（1+10%）
=160×1.1
=176
A: 176 pieces are processed every day

A car drives from the east to the West. After a journey, it is 210 kilometers away from the west, and then 20% of the whole journey. At this time, the ratio of the distance traveled to the distance not traveled is 3:2. How many kilometers are there between the East and the west?

Suppose the distance between the East and the west is x km, we can get: & nbsp; x-210 + 20% x = 35x, x-210 + 0.2x = 0.6x, & nbsp; 1.2x-210 = 0.6x, & nbsp; 1.2x-0.6x = 210, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 0.6x = 210, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 3

A car drives from the east to the West. After a journey, it is 210 kilometers away from the west, and then 20% of the whole journey. At this time, the ratio of the distance traveled to the distance not traveled is 3:2. How many kilometers are there between the East and the west?

Suppose the distance between the East and the west is x km, we can get: & nbsp; x-210 + 20% x = 35x, x-210 + 0.2x = 0.6x, & nbsp; 1.2x-210 = 0.6x, & nbsp; 1.2x-0.6x = 210, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 0.6x = 210, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 3

The distance between the two places is 210 km. How many km does a and B run from each other for 3.5 hours?
Column equation to explain. How to column, and so on

Set car B to travel x kilometers per hour
（28+x）×3.5=210
（28+x）=60
x=32
A: car B travels 32 kilometers per hour
28 + X is the sum of speed, and then multiplied by time, that's the total distance