The equation AX & # 178; + 2 (a-3x) + (A-2) = 0 of X has at least one integer solution, and a is an integer For example 4, the equation AX & # 178; + 2 (A-3) x + (A-2) = 0 has at least one integer solution, and a is an integer, and the value of a is obtained. When a = 0, the original equation becomes - 6x-2 = 0, and there is no integer solution When a ≠ 0, the equation is a quadratic equation with one variable, which has at least one integer root. It shows that the discriminant Δ = 4 (A-3) &# 178; - 4a (A-2) = 4 (9-4a) is a complete square number, so 9-4a is a complete square number. Let 9-4a = M & # 178;, then M is a positive odd number

The equation AX & # 178; + 2 (a-3x) + (A-2) = 0 of X has at least one integer solution, and a is an integer For example 4, the equation AX & # 178; + 2 (A-3) x + (A-2) = 0 has at least one integer solution, and a is an integer, and the value of a is obtained. When a = 0, the original equation becomes - 6x-2 = 0, and there is no integer solution When a ≠ 0, the equation is a quadratic equation with one variable, which has at least one integer root. It shows that the discriminant Δ = 4 (A-3) &# 178; - 4a (A-2) = 4 (9-4a) is a complete square number, so 9-4a is a complete square number. Let 9-4a = M & # 178;, then M is a positive odd number

First of all, we know that: the square of an even number is even, and the square of an odd number is odd. Because when a is an integer, 9-4a = 4 (1-A) + 5, so 9-4a is odd, that is M & # 178; is odd, and because the square of an odd number is odd, and the square of a number is odd, then the number m must be odd

If the equation x & # 178; - 3x + 1 = 0 is in the interval (2,3)
The number of roots

The number of roots is 1
The equation is symmetric with x = 3 / 2 and monotonically increasing in the interval (2,3)
When x = 2, X & # 178; - 3x + 1 = - 1; when x = 3, X & # 178; - 3x + 1 = 1
So there is one

The product of all roots of equation x2-3x-6 = 0 and equation x2-6x + 3 = 0 is______ ．

The product of two roots of equation x2-3x-6 = 0 is - 6, and the product of two roots of equation x2-6x + 3 = 0 is 3, so the product of all roots is - 18

Uncle Li takes a taxi to the airport 20 kilometers away. How much is the fare? Starting price: 6 yuan (within 5 kilometers, including 5 kilometers)
The excess distance is calculated as 1 yuan and 1 kilometer

20 is more than 5 km, so the fare should be 6 + (20-5) = 21 yuan

Increase the width of a rectangle by 5cm and decrease the length by 3cm, then it becomes a square. The area of a known square is 75cm square
The side length of the square is:
（75-3×5）÷（5-3）
=（75-15）÷2
=60÷2
=30 cm
The square area is:
30 × 30 = 900 square centimeter
There's a way to write that. I just want to ask,

I don't quite understand this solution. My solution is:
Let the side length of the square be x, then the length of the original rectangle is x + 3 (CM), and the width is X-5 (CM),
According to the meaning of the question, X2 = (x + 3) (X-5) + 75, the solution is x = 30
So the square area is: 30 × 30 = 900 square centimeters

In the triangle ABC, the corresponding sides of three internal angles a, B and C are a, B and C respectively. Given that C = 3, C = 60 degrees and a + B = 5, then the value of COS ((a-b) / 2) is?
RT

C=60°,A+B=120°,c=3,a+b=5,
cosC=(a²+b²-c²)/(2ab)=1/2,
It's time to tidy up
3ab+9=(a+b)²=25,
So AB = 16 / 3,
∴
cosA+cosB
=(b²+c²-a²)/(2bc)+(a²+c²-b²)/(2ac)
=[ab(a+b)+c²(a+b)-(a+b)(a²-ab+b²)]/(2abc)
=5/6,
∵cos[(A+B)/2]cos[(A-B)/2]
=(1/2)[cosA+cosB]
=5/12,
And COS [(a + b) / 2]
=cos60°
=1/2,
∴cos[(A-B)/2]=5/6,

In the following, fill in 1, 2, 3, 4, 5, 6, 7, 8 and 9 (the number in each formula cannot be repeated, and the numerator of fraction part is less than denominator), so that the value of formula a with fraction is the largest and that of formula B is the smallest

(1) 98745-1236, 12469 + 3578 and 12469 + 3578

Cut out the largest circle in a square paper with a side length of 50 cm. What is the area of this circle?

50÷2＝25cm25×25×3.14＝

If a squared + M = (a-half) (a + Half), then M=

(a-1/2)(a+1/2)=a²-1/4
m=-1/4

At the beginning of this year, a fishery company purchased a fishing boat with 980000 yuan for fishing. In the first year, it required 120000 yuan of various expenses. From the second year, the annual expenses including maintenance expenses increased by 40000 yuan compared with the previous year. The total fishing income of the fishing boat is expected to be 500000 yuan per year. (1) the fishing boat has made profits in several years (that is, the difference between the total accumulated income minus the cost and all expenses is positive)? (2) After fishing for several years, there are two treatment schemes: ① when the annual average profit reaches the maximum, sell the ship at 260000 yuan; ② when the total accumulated profit reaches the maximum, sell the ship at 80000 yuan? And explain the reason

1) Let the profit after n years be y yuan, then y = 50N − [12n + n (n − 1) 2 × 4] − 98 = − 2n2 + 40n − 98, so that Y > 0, 3 ≤ n ≤ 17, profit from the third year. 2) ① average profit yn = − 2n − 98n + 40 ≤− 22n × 98n + 40 = 12, in this case, the total profit is 12 × 7 + 26 = 1.1 million yuan, in this case, n = 7. ② y = - 2 (N-10) 2 + 102 ≤ 102, in this case, the profit is 102 + 8 = 1 10. The profit of both cases is 1.1 million yuan, the profit is the same, but the time of scheme 1 is short, so scheme 1 is cost-effective