Formula method factorization problem, urgent X & sup2; (x-3) + N + 3 power of 3-x-n-1 power of X 2 power of 1.222 * 9-2 power of 1.333 * 4 I know how to write,

Formula method factorization problem, urgent X & sup2; (x-3) + N + 3 power of 3-x-n-1 power of X 2 power of 1.222 * 9-2 power of 1.333 * 4 I know how to write,

X2 (x-3) + 3-x = X2 (x-3) - (x-3) = (x-3) (x2-1) = (x-3) (x + 1) (x-1) x ^ (n + 3) - x ^ (n-1) = x ^ (n-1 + 4) - x ^ (n-1) = x ^ (n-1) x ^ 4-x ^ (n-1) = x ^ (n-1) (x ^ 4-1) = x ^ (n-1) (x ^ 2 + 1) (x + 1) (x-1)
x²（x-3）+3-x =x²（x-3）-(x-3)=(x²-1)（x-3)=(x+1)(x-1)(x-3)
Solve the following equations by the method of collocation, formula or factorization,
x²+2（√2-1）x+3-2√2=0
In addition: √ = root
x²+2（√2-1）x+3-2√2=0
[x+(√2-1)]²x+3-2√2-(√2-1)²=0
[x+(√2-1)]²=0
x=1-√2
It is known that the quadratic equation AX ^ 2 + BX + C = 0 with real coefficients of X has two imaginary roots X1 and x2. If | x1-x2 | = 2 and 2 + AI = C-1 + I, find the roots X1 and X of the equation
2 + AI = C-1 + I = = > A = 1, C = 3, | x1-x2 | = 2 = = > radical (4ac-b ^ 2) / a = 2, B ^ 2 = 8, B = ± 2, radical 2
X1,2 = radical 2 ± I or - radical 2 ± I
By analogy with factorization, a quadratic equation of one variable with X as unknowns and - 2 and 4 as roots is written
x=-2
x+2=0
X=4
x-4=0
So (x + 2) (x-4) = 0
x²-2x-8=0
In the polynomial: 1 / 2 x square + X-1, 1 / 2 x square = 3x + 1, x square - x, choose any two to add and decompose the result into factors
Choose 1 / 2x ^ 2 + X-1, x ^ 2-x
1/2x^2+x-1+x^2-x
=3/2x^2-1
=(3/2x+1)(3/2x-1)
How to solve a math problem in Junior Three
The fourth power of 3 * y + the third power of 2 * y - 1 = 0 (* is a multiplication sign)
3y^4+2y³-1=0；
3y^4+3y³-y³-1=0；
3y³（y+1）-（y³+1）=0；
3y³（y+1）-（y+1）（y²+1-y）=0；
（y+1）（3y³-y²+y-1）=0；
∧ y = - 1 or 3Y & sup3; - Y & sup2; + Y-1 = 0, the following is difficult to solve by using junior high school knowledge
ax²-16ay²
ax²-16ay²
=a(x²-16y²)
=a(x+4y)(x-4y)
1： a(x²-16y²)
2： a(x+4y)(x-4y)
=A (x-16y) = a (x + 4Y) (x-4y) square, I can't type, use words
Find a polynomial that is 4-2 / 1 x ^ 2 less than 3x ^ 2-4x-5, and find the value of this polynomial when x = 2
3x ^ 2-4x-5 - (1 x ^ 2 of 4-2)
=（7/2）x^2-4x-9
Let x = 2
Then (7 / 2) x ^ 2-4x-9 = 14-8-9 = - 3
(x-1) (X-2) (x-3) (x-4) = 0,
(x-1)(x-2)(x-3)(x-4)=0
(x-1)=0
perhaps
(x-2)=0
perhaps
(x-3)=0
perhaps
(x-4)=0
The solution is x = 1 or x = 2 or x = 3 or x = 4
X=1,2,3,4
To be honest, you can see the answer by looking at it without writing the process x = 1 or x = 2 or x = 3 or x = 4
Factorization: 1-x2 + 4xy-4y2=______ .
The original formula is 1 - (x2-4xy + 4y2) = 1 - (x-2y) 2 = (1 + x-2y) (1-x + 2Y), so the answer is (1 + x-2y) (1-x + 2Y)