Given - x + 2Y = 5, then the value of 5 (x-2y) 2-3 (x-2y) - 60 is () A. 80B. 10C. 210D. 40

Given - x + 2Y = 5, then the value of 5 (x-2y) 2-3 (x-2y) - 60 is () A. 80B. 10C. 210D. 40

∵ - x + 2Y = 5, ∵ x-2y = - 5, the original formula = 5 (x-2y) 2-3 (x-2y) - 60 = 5 × (- 5) 2-3 × (- 5) - 60 = 125 + 15-60 = 80

If the square of (X-2) + = 0 (absolute value), find the cube of minus 2x minus the square of minus y. The answer is minus 41,

The sum of squares of (X-2) is greater than or equal to zero. If their sum is zero, it means that they are zero respectively. Therefore, x = 2, y = - 5. Therefore, the square of - 2x Cube - (- y) = - 41

The rule of "24 o'clock game" is as follows: take any four numbers and perform four operations of addition, subtraction, multiplication and division on these four numbers (each number can only be used once), so that the result is 24. There are four numbers 3, - 5,7 and - 13. Use the above rules to write an operation formula, so that the result is 24 (brackets can be used). The operation formula you designed is____________ .

[(-5)×(-13)+7]÷3=24

（-52）+（-19）-（+37）-（-24）

The original formula = - 52-19-37 + 24 = - 108 + 24 = - 84

The two sides of the equilateral triangle ABC are respectively extended to 2 times of the original, and a new triangle is obtained. How many times of the area of the new triangle is that of the original equilateral triangle?
If each side of the triangle ABC is extended to twice of the original, a new triangle is obtained. How many times is the area of the new triangle?

4 times

19.5x-3 = 12

19.5x=15 x=10/13

A parallelogram has the same base and height as a triangle. If the area of the triangle is 256dm, the area of the parallelogram is, if the bottom of the triangle is 16dm, the height is

512 32
256X2=512
256x2 divided by 16 = 32

1/1997*1998+1/1998*1999+1/1999*2000+.+1/2005*2006+1/2006*2007+1/2007=?

Because 1 / N (n + 1) = 1 / n-1 / (n + 1)
So the original formula = 1 / 1997-1 / 1998 + 1 / 1998-1 / 1999 +... + 1 / 2006-1 / 2007 + 1 / 2007
=1/1997

As shown in the figure, we know that D and E are the points on the sides of AB and AC of △ ABC respectively, and △ ABC ∽ ade, ad: DB = 1:3, de = 2, find the length of BC

∵ ad: DB = 1:3, ∵ ad: ab = 1:4, (2 points) ∵ ABC ∽ ade ∵ ad: ab = de: BC (2 points) ∵ de = 2 ∵ BC = 8. (2 points)

Given a + B + C = 1, a (1 / B + 1 / C) + B (1 / A + 1 / C) + C (1 / A + 1 / b) = - 3, find a + B + C
Come on, today~~

A (1 / B + 1 / C) + B (1 / A + 1 / C) + C (1 / A + 1 / b) = - 3
[a (1 / B + 1 / C) + 1] + [b (1 / A + 1 / C) + 1] + [C (1 / A + 1 / b) + 1] = 0
A (1 / A + 1 / B + 1 / C) + B (1 / A + 1 / B + 1 / C) + C (1 / A + 1 / B + 1 / C) = 0
(a + B + C) (1 / A + 1 / B + 1 / C) = 0
(1) If a + B + C = 0, the answer is 0
(2) 1 of a + 1 of B + 1 of C = 0
Then (AB + BC + Ca) = 0
So AB + BC + Ca = 0
And because a & # 178; + B & # 178; + C & # 178; = 1
So a & # 178; + B & # 178; + C & # 178; + 2Ab + 2BC + 2ca = 1
(a+b+c)²=1
So a + B + C = ± 1
In conclusion, a + B + C = 0 or ± 1