Given X & # 178; - X-1 = 0, what is the value of - X & # 179; + 2x & # 178; + 2013?

Given X & # 178; - X-1 = 0, what is the value of - X & # 179; + 2x & # 178; + 2013?

It is known that X & # 178; - X-1 = 0,
x²=x+1
Then - X & # 179; + 2x & # 178; + 2013
=-x(x+1)+2(x+1)+2013
=-x²-x+2x+2+2013
=-(x+1)+x+2015
=-x-1+x+2015
=2014
X & # 178; = x + 1, so x & # 179; = x & # 178; + X
-x³+2x²+2013=-x²-x+2x²+2013=x²-x+2013=x+1-x+2013=2014.
[countless] the Legion will answer for you. If you don't understand the question, you can choose to adopt it^_^
May I ask the solution formula of higher order equation, for example, the solution formula of degree 4 or 5
Dichotomy, after deriving, find the value space, and then divide,
Can (1-x) ^ 2-4 (1-x ^ 2) + 4 (1 + x) ^ be factorized by the complete square formula?
The Ball..
By the way, solve another problem: the square of (a + b) - the square of 4A, the square of B
One
(1-x)²-4(1-x²)+4(1+x)²
=(1-x)²-2*(1-x)*[2(1+x)]+[2(1+x)]²
=[(1-x)-2(1+x)]²
=(3x+1)²
Two
(a+b)²-4a²b²
=(a+b+2ab)(a+b-2ab)
When x = 2013, (X & # 178; - 2x + 1) &# 178; △ (x-1) &# 179;
1/2013
The formula method of quadratic equation with one variable in junior three mathematics
Given that real numbers x and y satisfy 20x ^ 2-12xy + 2Y ^ 2-8x + 2Y + 1 = 0, find the value of X and y
Try to solve the problem by formula
2y^2+(2-12x)y+(20x^2-8x+1)=0
Discriminant = (2-12x) ^ 2-8 (20x ^ 2-8x + 1)
=-4(4x^2-4x+1)
=-4(2x-1)^2
If 2x-1 ≠ 0
Then the discriminant is less than 0 and there is no solution
So x = 1 / 2
2y^2-4y+2=0
2(y-1)^2=0
So x = 1 / 2, y = 1
1. The square of a minus the square of B plus B minus one fourth, we need to use the factorization of grade one,
2.4 (square of x plus y) (square of x minus y) minus (square of 2x minus y) square
Where x = 2Y = - 5
First simplify, in the evaluation, the brackets inside are the brackets in the title, which I need urgently
A²-B²+B-1/4
=A²-(B²-B+1/4)
=A²-(B-1/2)²
=(A+B-1/2)(A-B+1/2)
4(X²+Y)(X²-Y)-(2X²-Y)²
=4(X^4-Y²)-(4X^4-4X²Y+Y²)
=4X^4-4Y²-4X^4+4X²Y-Y²
=4X²Y-5Y²
=4*4*(-5)-5*25
=-205
1.A2-B2+B-1/4=A2-(B2-B+1/4)=A2-(B-1/2)2=(A+B-1/2)(A-B+1/2)
2.4(X2+Y)(X2-Y)-(2X2-Y)2=4(X4-Y2)-(4X4-4X2*Y+Y2)=-5Y2+4X2*Y
Just put in the numerical calculation
1)a^2-b^2+b-1/4
=a^2-(b^2-b+1/4)
=a^2-(b-1/2)^2
=(a+b-1/2)(a-b+1/2)
Complete square formula
2）4（x^2+y）(x^2-y)-(2x^2-y)^2
=4(x^4-y^2)-(4x^4-4x^2y+y^2)
=-4y^2+4x^2y-y^2
=4x^2y-5y^2
=100*(-5/2)-25/4
=-250-25/4
=-1025/4
(1)a²-b²+b-1/4==A²-(B²-B+1/4)
=A²-(B-1/2)²
=(A+B-1/2)(A-B+1/2）
（2）=4（x²＋y）(x²-y)-(2x²-y)²
= 4x^4-4y²-(2x²-y)²
=... unfold
(1)a²-b²+b-1/4==A²-(B²-B+1/4)
=A²-(B-1/2)²
=(A+B-1/2)(A-B+1/2）
（2）=4（x²＋y）(x²-y)-(2x²-y)²
= 4x^4-4y²-(2x²-y)²
=(2x²)²-(2x²-y)²-4y²
=(2x²+2x²-y)(2x²-2x²+y)-4y²
=(4x²-y)y-4y²
=y(4x²-5y)
=-5×(16+25)
=-205. Put it away
2X & # 178; + 3x + 1 = 0 to find x
X = – 1 or – 0.5
It is proved that no matter what the value of M is, the equation 2x2 - (4m-1) x-m2-m = 0 always has two unequal real roots
It is proved that: ∵ △ = [- (4m-1)] 2-4 × 2 × (- m2-m) = 24m2 + 1 ＞ 0 ∵ has two unequal real roots
1:
(x+1)(x+3)(x+5)(x+7)+15
2:
x^2-2xy+y^2-13x+13y+30
3:
Let's decompose the factor (y = 2x / 3, y = 2x / 3) + 2
4:
It is known that a, B and C are three sides of triangle ABC, and a, B and C satisfy (a + B + C) ^ 2 = 3 (a ^ 2 + B ^ 2 + C ^ 2)?
One
(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x+7)[(x^2+8x+7)+8]+15
=(x^2+8x+7)^2+8(x^2+8x+7)+15
=[(x^2+8x+7)+3][(x^2+8x+7)+5]
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)
Two
x^2-2xy+y^2-13x+13y+30
=(x-y)^2-13(x-y)+30
=(x-y-3)(x-y-10)
3:
First, we decompose the factor 4 (x + 2Z) (x-2z) + 12xy + 9y ^ 2, where x = 1 / 2, y = 2 / 3, z = 3 / 4
Original formula = 4 (x ^ 2-4z ^ 2) + 12xy + 9y ^ 2 = (4x ^ 2 + 12xy + 9y ^ 2) - 16Z ^ 2
=(2x+3y)^2-16z^2
=(2x+3y+4z)(2x+3y-4z)
=(1+2+3)(1+2-3)
=0
4:
It is known that a, B and C are three sides of triangle ABC, and a, B and C satisfy (a + B + C) ^ 2 = 3 (a ^ 2 + B ^ 2 + C ^ 2)?
(A+B+C)^2=3(A^2+B^2+C^2)
A^2+B^2+C^2+2AB+2AC+2BC=3A^2+3B^2+3C^2
2AB+2AC+2BC=2A^2+2B^2+2C^2
A^2-2AB+B^2+A^2-2AC+C^2+B^2-2BC+C^2=0
(A-B)^2+(A-C)^2+(B-C)^2=0
A-B=0,A-C=0,B-C=0
A=B=C
Equilateral triangle
One
(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x+7)[(x^2+8x+7)+8]+15
=(x^2+8x+7)^2+8(x^2+8x+7)+15
=[(x^2+8x+7)+3][(x^2+8x+7)+5]
Solve the equation, &; √ 3x &; - X &; - 2 √ 3 = 0
Note: 3x & # 178; the one on the left and the second 3, the one on the left is the root
Change √ 3x & # 178; - X-2 √ 3 = 0