If X & # 178; - Y & # 178; - x + y = (X-Y) * a, then a=__________ . If X & # 178; - Y & # 178; - x + y = (X-Y) * a, then a=__________ .

If X & # 178; - Y & # 178; - x + y = (X-Y) * a, then a=__________ . If X & # 178; - Y & # 178; - x + y = (X-Y) * a, then a=__________ .

A:
x^2-y^2-x+y
=(x-y)(x+y)-(x-y)
=(x-y)(x+y-1)
=(x-y)*A
So: a = x + Y-1

X-Y = 1, find X & # 178; - Y & # 178; - 2Y

x-y=1
So the original formula = (x + y) (X-Y) - 2Y
=(x+y)*1-2y
=x+y-2y
=x-y
=1

If the univariate quadratic equation (M-3) x ^ 2-x + 9-m ^ 2 = 0 has a root of 0, find the value of M

Take x = 0 into the equation
We get: 9-m & # 178; = 0, M = 3 or - 3
But because it is a quadratic equation of one variable, M-3 is not equal to 0, M is not equal to 3
So m = - 3
Hope to adopt!

It is known that - 2 is a root of quadratic x ^ 2 + MX + n = 0 of one variable, and the value of the root number 8m ^ 2-8mn + 2n ^ 2 of quadratic radical formula is obtained

Take - 2 into the equation to get 4-2m + n = 0 and N = 2m-4
√（8m^2-8mn+2n^2）
=√[8m^2-8m（2m-4）+2(2m-4)^2]
=√(8m^2-16m^2+32m+8m^2-32m+32)
=√32
=4√2

"M ＜ 14" is the () condition of "quadratic equation with one variable & nbsp; x 2 + X + M = 0 has real number solution"
A. Sufficient but not necessary B. sufficient and necessary C. necessary but not sufficient D. neither sufficient nor necessary

In order to make the quadratic equation with one variable have real number solution, the discriminant △ = 1-4m ≥ 0, and the solution is m ≤ 14. So "m < 14" is a sufficient and unnecessary condition for the quadratic equation with one variable to have real number solution

It is known that the reciprocal sum of the two unequal real roots of the quadratic equation m ^ 2x ^ 2 + 1 = (3-2m) x is s, and the value range of S is obtained

M & sup2; X & sup2; + 1 = (3-2m) XM & sup2; X & sup2; - (3-2m) x + 1 = 0 the equation has two unequal real roots, then the discriminant is greater than 0 Δ = [- (3-2m)] & sup2; - 4m & sup2; = - 12m + 9 > 0, m < 3 / 4 and m ≠ 0. Suppose that the two real roots of the equation are X1 and X2, and by Weida's theorem, X1 + x2 = (3-2m) / M & sup2;, x1x2 =

Let 2005x ^ 3 = 2006y ^ 3 = 2007z ^ 3, XYZ > 0, and 3 ^ √ (2005x ^ 2 + 2006y ^ 2 + 2007z ^ 2) = 3 ^ √ 2005 + 3 ^ √ 2006 + 3 ^ √ 2007, find 1 / X+
1 / y + 1 / Z

Hello
set up
2005x^3=2006y^3=2007z^3=t
2005x^2=t/x
2006y^2=t/y
2007z^2=t/z
³√2005=（³√t）/x
³√2006=（³√t）/y
³√2007=（³√t）/z
From the known
³√（t/x+t/y+t/z）=（³√t）/x+（³√t）/y+（³√t）/z=³√t（1/x+1/y+1/z）
The third power of both sides
t/x+t/y+t/z=t（1/x+1/y+1/z）＾3
1/x+1/y+1/z=（1/x+1/y+1/z）＾3
（1/x+1/y+1/z）＾2=1
xyz>0,
From the condition, we get that x, y and Z have the same sign
x>0,y>0,z>0,
1/x+1/y+1/z>0,
1/x+1/y+1/z=1
If you don't understand, please ask. If you understand, please select it as a satisfactory answer in time! (*^__ ^*Thank you!

A rectangular board, 18 decimeters long, 14 decimeters wide, if it is made into a largest round desktop, minus the area of the board is more than square decimeters

gykmhjjn,
The area of the round table is:
3.14 × (14 △ 2) ^ 2 = 153.86 (square decimeter)
The area of the board is:
18 × 14 = 252 (square decimeter)
The area subtracted is:
252-153.86 = 98.14 (square decimeter)

Saw the biggest circle from a 10 decimeter long and 8 decimeter long rectangular board. The area of the circle is () square decimeter

Length 10 decimeters, width 8 decimeters? That is: width is diameter divided by 8, 2 = 4 decimeters (radius) and then 3.14 times 4 square = 50.24 answer: the area of a circle is (50.24) square decimeters. Rest assured, I am good at mathematics

On a piece of rectangular paper 6 decimeters long and 4 decimeters wide, after cutting the largest circle, what is the remaining area of cubic decimeters?

6 × 4-3.14 × (4 △ 2) 2 = 24-12.56 = 11.44 square decimeters. A: the remaining area is 11.44 square decimeters