(it's better to understand. X is the death of Ike and X is the multiply sign) judge 1. 16x = 0 is the equation () 2. The number 5 less than x is 10, and X is 10 () 3. When x is reduced by 25%, it is equal to 3 / 4 of 300. X is 300 () 4. When solving the equation 3x + 5 / 4 × 2 = 13 / 3, we should first consider 3x as an additive () 5. A car travels an average of M kilometers per hour, a bicycle travels an average of N kilometers per hour (n < m), and the distance difference is 4 (m-n) kilometers () solve equations x+48=72 2x-5=25 0.15x=0.9 9÷5x=12 4x+36=56 10=31-3x 1.7×(12-x)=5.1 (135 + 49) △ x = 23.4% 1 (x + 0.25) = 45% Practical questions 1. The school chorus has 124 people, 19 more than three times the number of science and technology groups. How many people are there in the school science and technology groups? 2. After 2.5 hours of driving on a 360 km long road, there are 210 km left. How many kilometers does the car travel per hour?

(it's better to understand. X is the death of Ike and X is the multiply sign) judge 1. 16x = 0 is the equation () 2. The number 5 less than x is 10, and X is 10 () 3. When x is reduced by 25%, it is equal to 3 / 4 of 300. X is 300 () 4. When solving the equation 3x + 5 / 4 × 2 = 13 / 3, we should first consider 3x as an additive () 5. A car travels an average of M kilometers per hour, a bicycle travels an average of N kilometers per hour (n < m), and the distance difference is 4 (m-n) kilometers () solve equations x+48=72 2x-5=25 0.15x=0.9 9÷5x=12 4x+36=56 10=31-3x 1.7×(12-x)=5.1 (135 + 49) △ x = 23.4% 1 (x + 0.25) = 45% Practical questions 1. The school chorus has 124 people, 19 more than three times the number of science and technology groups. How many people are there in the school science and technology groups? 2. After 2.5 hours of driving on a 360 km long road, there are 210 km left. How many kilometers does the car travel per hour?

Practical questions:
1. Set up the school science and technology group, there are x people
3x+19=124
2. Suppose the car travels an average of X kilometers per hour
2.5x+210=360
solve equations:
x+48=72 2x-5=25 0.15x=0.9 9/5x=12 4x+36=56 10=31-3x
x=72-48 2x=25+5 x=0.9÷0.15 5x=12×5 4x=56-36 3x=31-10
x=? 2x=30 x=? 5x=60 4x=20 3x=11
x=30÷2 x=60÷5 x=20÷4 x=11÷3
x=? x=? x=? x=?
1.7×（12-x）=5.1 （135+45）÷x=23 1/4×（x+0.25）=45%
12-x = 1.7 × 5.1 180 △ x = 23 / 4 x + 0.25 = 0.45
12-x = 8.67 x = 180 △ 23 1 / 4 x = 0.45-0.25
X = 12-8.67 x =? 1 / 4 x = 0.2
X =? X = 0.2 △ 1 / 4
x=?
Judgment questions:
1 pair
2 wrong
Three mistakes
4 pairs
5 pairs

Given that the vector ABC satisfies | a | = | B | = 2, | C | = 1, (A-C) (B-C) = 0, then the value range of | A-B | is?

(a-c)(b-c)=0---->ab-c(a+b)+c^2=0---->(ab+1)^2=[c(a+b)]^2,
(ab)^2+2(ab)+1

As shown in the figure, it is known that in the triangle ABC, the bisectors AD and be of ∠ C = 90 °, BAC and ABC intersect point O, and the degree of ∠ AOB is calculated

∠ AOB = 180 ° - (∠ a + ∠ b) * half = 135 °

After the improvement of corn varieties, the average yield of corn in Yingchun village increased by a ton per hectare. The total yield of a field that used to produce M tons of corn has increased by 20 tons. What is the average yield of corn per hectare?

Suppose that the average yield of corn per hectare is x tons, then the average yield of corn per hectare is (x + a) tons. From the problem meaning, there is MX = m + 20x + A, the solution is x = mA20

Let F 1 and F 2 be the two focal points on the ellipse x ^ 2 / 4 + y ^ 2 = 1, p be on the ellipse, when the area of △ f 1pf 2 is 1, the vector Pf1 · vector PF2 =?

Let P (x, y), because a ^ 2 = 4, B ^ 2 = 1, so C ^ 2 = a ^ 2-B ^ 2 = 3, because sf1pf2 | * | y | = √ 3 | y | = 1, so | y | = 1 / √ 3, substituting into the elliptic equation, we can get x ^ 2 = 8 / 3, so Pf1 * PF2 = (- √ 3-x, - y) * (√ 3-x, - y) = x ^ 2 + y ^ 2-3 = 8 / 3 + 1 / 3-3 = 0

It is known that: as shown in the figure, in the equilateral triangle ABC, D and E are the points on BC and AC respectively, and AE = CD, connect AD and be to point P, make BQ ⊥ ad, and the perpendicular foot is Q. verification: BP = 2pq

It is proved that: ∵ ABC is an equilateral triangle, ∵ AB = AC = BC, ∵ C = ABC = 60 °, ∵ AE = CD, ∵ EC = BD; ≌ BEC ≌ ADB (SAS), ∵ EBC = ∵ bad; ∵ Abe + ∩ EBC = 60 °, then ∫ Abe + ∩ bad = 60 °, and ∫ bpq is the outer angle of △ ABP, ∫ ABP + ≌ BAP = 60 ° = ≌ bpq, and