The square of 2x, the cube of y-4x, the square of y-6xy

The square of 2x, the cube of y-4x, the square of y-6xy

2x^2y^3-4x^3y^2-6xy
=2xy(xy^2-2x^2y-3)

Factorization factor of 2x cube y + 4x square y-6xy cube

2x^3y+4x^2y^2-6xy^3
=2xy(x^2+2xy-3y^2)
=2xy(x+3y)(x-y)

Multiplication and division of fractions 1 - (square of 2 ／ 4x, square y) ／ cubic power of 2 ／ 5xy

Should there be a bracket after the second division sign
- (the square of 2 △ 4x, the square of Y) / (the third power of 2 △ 5xy)
=-(1/x^2*y)/(27/5xy)
=-5/(27x)

Range x2 + 2x-1 (- 2)

X2 + 2x-1 = x2 + 2x + 1-2 = (x + 1) square-2, according to this formula, we can see that this is a parabola with (- 1, - 2) as the vertex and x = - 1 as the axis of symmetry

Find the range of function y = - x2 + 2x + 5 (- 1 less than or equal to x less than or equal to 4)

y=(x+2)^2+1
The range is (2)

y=x2-2x+5(-1

y=x²-2x+1+4=(x-1)²+4
The opening is upward, and the axis of symmetry x = 1
So x = 1, the minimum is 4
X = - 1 or 3, up to 8
So the range is [4,8]

Find the range y = x2-2x-3 / x2-5x-6

It is recommended to use Δ discriminant method: y = (X & # 178; - 2x-3) / (X & # 178; - 5x-6) x & # 178; - 2x-3 = x & # 178; y-5xy-6y (1-y) x & # 178; + (5y-2) x + (6y-3) = 0, because the equation about X has solution; so (5y-2) &# 178; - 4 (1-y) (6y-3) ≥ 049y & # 178; - 56Y + 16 ≥ 0 (7y-4) &# 178; ≥ 0y ∈ R

The function y = x2-2x = 3, X belongs to {- 1,0,1,2}, and the range is

When x = - 1, y = 0
When x = 0, y = - 3
When x = 1, y = - 4
When x = 2, y = - 3
So the range of Y is - 3,0, - 4

(2008-A) (2009-A) = 1999, then (2008-A) ^ 2 + (2009-A) ^ 2=
Like the title,

(2008-a)^2+(2009-a)^2=(2008-a-2009+a)^2+2(2008-a)(2009-a)
=1+2*1999=3999

98 times 2 of 99 = clever calculation 2009 times 2008 times 2008

98*2/99
=（99-1）*2/99
=2-2/99
=1 and 97 / 99
2008/2009*2008
=2008/2009*（2009-1）
=2008-2008/2009
=January 2007 / 2009