Two math problems (write the calculation process) 1.（-5x+3/2）（-5x－1.5） 2.（x+2y－3）（x－2y+3）

Two math problems (write the calculation process) 1.（-5x+3/2）（-5x－1.5） 2.（x+2y－3）（x－2y+3）

So: 1, (- 5x + 3 / 2) (- 5x-1.5) = (- 5x) * (- 5x) - 1.5 * 1.5 = 25X * x-2.252, (x + 2y-3) (x - (2y-3)) = x * x - (2y-3) * (2y-3) = x * x-4y * y + 12y-9

1. Triangle ABC is an equilateral triangle with side length of 1, triangle BDC is an isosceles triangle with vertex angle BDC of 120 degrees, and make an angle of 60 degrees with D as vertex. The two sides of the angle intersect AB and m respectively, and AC intersects n to connect Mn
Verification: the perimeter of triangle amn is 2
2. Take sides AB and BC of triangle ABC as sides, make square abmn and bcpq outside the triangle respectively, the midpoint of AC is D, and the straight line BD intersects MQ at E
Verification: the straight line BD is perpendicular to MQ

1. Extend MB to e so that be = CN because of equilateral triangle ABC, angle ABC = angle ACB = 60 degrees because triangle BDC is an isosceles triangle with vertex BDC of 120 degrees, so angle CBD = angle BCD = 30 degrees, DB = DC, so angle ABC + angle CBD = angle ACB + angle BCD = 90 degrees, so angle abd = angle DBE = angle ACD = 90 degrees, because

If a square + AB = 15. B square + AB = B, find the value of the sum of a square and B squares (a + b)

From B + AB = B, put forward B, then B (B + a) = B, so (a + b) = 1, a + AB = 15, put forward a, then a (a + b) = 15, because a + B = 1, so a = 15, so B = - 14, so A-B = 29

The derivative f '(x) = KX + B of the function f (x) defined on R, where the constant k > 0, then the function f (x) is
Increasing on a (- infinity, + infinity), increasing on B [- B / K, + infinity]
Increasing on C (- infinity, - B / k) and decreasing on D (- infinity, + infinity)

Let f '(x) > 0, then KX + b > 0
‖ KX > - B, x > - B / K (k > 0, ‖ inequality invariant sign)
That is, when x > - B / K, f '(x) > 0, then f (x) increases singly
Choose B
Do not understand can ask, help please adopt, thank you!

Find a math problem
Known: specify the perimeter and area of a two-dimensional figure,
And the shape of this figure is arbitrary
Q: how many graphics meet the requirements?

Building owner,
The answer is countless!
Let's take a piece of square paper. You subtract a corner and put it in other places (the joint is intact). The area and perimeter are the same,

If the quadratic function y = x2-2x + 3 is changed into the form of y = (X-H) 2 + K, then H + K=______ ．

∵ y = x2-2x + 3 = x2-2x + 1 + 2 = (x-1) 2 + 2, ∵ H = 1, k = 2, ∵ H + k = 1 + 2 = 3

The derivative of the function y = x ^ 2sinx is

y'=2xsinx+x^2cosx

50 exercises of decimal point addition and subtraction

1.0+1.5 21.2+1.5546.+21 54+.151.0+1.5 21.2+1.5546.+21 54+.151.0+1.5 21.2+1.5546.+21 54+.151.0+1.5 21.2+1.5546.+21 54+.151.0+1.5 21.2+1.5546.+21 54+.151.0+1.5 21.2+1.5546.+21 54+.151.0+1.5 21.2+1.5546....

Find the minimum m (a) and maximum m (a) of the function f (x) = x & sup2 - (2 + 6A & sup2) x + 3A & sup2 in the interval [0,1]

When a = 0, G (a) = - 3
When a ∈ (1 / 4,1 / 2], G (a) = 2a-1 / 4a-1
When a ∈ (1 / 2, + ∞), G (a) = 3a-2
When a ∈ (0,1 / 4), G (a) = 6a-3

It is known that the solution of the system of equations {ax + 5Y = 15, 1, 4x by = - 2, 2} is x = 5, y = 4
Given the equations {ax + 5Y = 15 ①, 4x by = - 2 ②}, because a misread a in equation ①, the solution of the equations is x = - 3, y = - 1. B misread B in equation ②, the solution of the equations is x = 5, y = 4. Try to calculate the value of a to the power of 2009 (- 10 / 1b) to the power of 2008

A misread the equation (1) and get the solution of the equations as x = - 3, y = - 1. Substituting, - 3a-5 = 15 to get a = - 20 / 3. B misread the equation (2) and get the solution of the equations as x = 5, y = 4 substituting into the equation (2), 20-4y = - 2 to get b = 11 / 2. So the equations {- 20x / 3 + 5Y = 15 (1), 4x-11y / 2 = - 2 (2)), simplify: 4x-3y = 9