Let x = the absolute value of a / (B + C) + B / (c + a) + C A + B, let ABC not be 0, and a + B + C = 0 The 19th power of x-99x + 2002

Let x = the absolute value of a / (B + C) + B / (c + a) + C A + B, let ABC not be 0, and a + B + C = 0 The 19th power of x-99x + 2002


A:
Let X be the absolute value of | a | / (B + C) + B / (c + a) + C / (a + b)
So:
x=| |a|/(-a)+|b|/(-b)+|c|/(-c)| |
1) If a is positive, B and C are negative
x=|-1+1+1|=1
2) If a and B are positive, C is negative
x=|-1-1+1|=|-1 |=1
To sum up, x = 1
So:
x^19-99x+2002
=1-99+2002
=2004-100
=1994



Given that a and B are rational numbers, compare the size of (a + b) and (a-b),


b> The former is greater than the latter
b



Given that a and B are positive rational numbers, let m = B / A, n = 2A + B / A + B, compare the size of M and n
Would you like to discuss the size of b square and 2A square?


[would you like to discuss the size of square B and square 2A
It can be said that, in fact, it is to discuss the proportional relationship between a and B
m=b/a
n=(2a+b)/(a+b)
m-n=b/a-(2a+b)/(a+b)=[b(a+b)-a(2a+b)]/[a(a+b)]
=(ab+b^2-2a^2-ab)/[a(a+b)]
=(b^2-2a^2)/[a(a+b)]
When B > (radical 2) a, m > n
When B = (radical 2) a, M = n
When B < (radical 2) a, m < n



Given that the parabola and X-axis intersect at points m (- 1,0) and n (2,0) and pass through points (1,2), find the expression of this function


Solution 1: substituting (- 1, 0), (2, 0), (1, 2) into y = AX2 + BX + C, we get a − B + C = 04A + 2B + C = 0A + B + C = 2, we get a = − 1b = 1C = 2, the expression of this function is y = - x2 + X + 2; solution 2: let the analytic expression of the function be y = a (x + 1) (X-2), substituting (1, 2) into y = - (x + 1) (X-2), that is y = - x2 + x + 2



Find the value of 1 / (x + 4) - 1 / (x + 7) = 1 / (x + 3) - 1 / (x + 6)


1/(x+3)-1/(x+4)=1/(x+6)-1/(x+7)
(x+6)(x+7)=(x+3)(x+4)
6x=-30
x=-5



The negative - x + 2Y is 3x-y, which does not change the value of the fraction, so that the first term of the molecule is positive, and the fraction itself does not have a "-" sign


Negative - x + 2Y / 3x-y
=3x-y of x-2y



Solving one variable quadratic equation 2x-2x2-2 / 1 ≥ 0


2x-2x & # 178; - 1 / 2 ≥ 0 is an equation or an inequality? 4x & # 178; - 4x + 1 ≤ 0 (2x-1) & # 178; ≤ 02x-1 = 0x = 1 / 2, so the solution of the original inequality is x = 1 / 2, the solution of the quadratic equation with one variable is 2x-2x2-2 / 1 = 04x & # 178; - 4x + 1 = 0 (2x-1) & # 178; = 02x-1 = 0x = 1 / 2, so the solution of the quadratic equation with one variable is x = 1 / 2



Fraction calculation: 1. (4 / m ^ 2-4) - (1 / m-2) 2. X + Y - (x ^ 2 + y ^ 2 / x + y) 3. 1 - (x multiplied by 1 / 1-x)


(4/m^2-4)-(1/m-2)
=4/(m+2)(m-2) -1/(m-2)
=4/(m+2)(m-2)-(m+2)/(m+2)(m-2)
=(4-m-2)/(m+2)(m-2)
=(2-m)/(m+2)(m-2)
=-1/(m+2)
2.x+y-(x^2+y^2/x+y)
=(x+y)²/(x+y)-(x²+y²)/(x+y)
=[(x+y)²-x²-y²]/(x+y)
=2xy/(x+y)
3.1 - (x multiplied by 1 / 1-x)
=1+x/(x-1)
=(x-1+x)/(x-1)
=(2x-1)/(x-1)



4(x+2)=1-5x 3(y+2)=3-2 y/2+x/3=13 y/3-x/4=3 x+y=420 30%x+40%y=160*180% 2x-1/5+3y-2/4=2 3x+1/3-3y+2/
4(x+2)=1-5x
3(y+2)=3-2
y/2+x/3=13
y/3-x/4=3
x+y=420
30%x+40%y=160*180%
2x-1/5+3y-2/4=2
3x+1/3-3y+2/4=4/3
x/2+y/3=2
2x+3y=28


4X + 8 = 1-5x (4 + 5) x = 1-8x = - 7 / 93y = 1-6y = - 5 / 3; Y / 2 = 13-x / 3Y = 26-2x / 3 (26-2x / 3) / 3-x / 4 = 3 - (2 / 9 + 1 / 4) x = 3-26 / 3x = 12Y = 26-24 / 3 = 18; y = 420-x0.3x + 0.4 (420-x) = 160 * 1.8-0.1x = 120x = - 1200y = 1620; by simplification, 2x + 3Y = 2.7 X-Y = 1 / 6x = y + 1 / 62 (y + 1



Use the reduced order method to calculate the following determinants
 


(1) r1-r2
Then c2-c1
Determinant into
x 0 0 0
1 -x 1 1
1 0 1+y 1
1 0 1 1-y
Expand by line 1 (reduced)
Then expand by column 1
(3)
Expand d = x * x ^ (n-1) + (- 1) ^ (n + 1) y * y ^ (n-1) = x ^ n + (- 1) ^ (n + 1) y ^ n by column 1