Factorization of the eighth grade ① - x * (- x) & # 179; * (- x) quartic power * (- x) & # 179; ② (- a) the 2n-3 power of quintic power * a - the 2n power of a * (- a) & # 178;

Factorization of the eighth grade ① - x * (- x) & # 179; * (- x) quartic power * (- x) & # 179; ② (- a) the 2n-3 power of quintic power * a - the 2n power of a * (- a) & # 178;

Is that a little star or a multiply sign
1. If it's an unknown number: - x (8 * + 3). The brackets are all power squares. I'm just a freshman. I don't know how to do it. Let's make do with it
2. - A (2n + 4) in brackets is the power
1：（a＋b） 2;＝a 2;＋2ab＋b 2; 2: （a－b） 2;＝a 2;－2ab＋b 2; 3: a 2;－b 2;＝（a－b）（a＋b） 4: （a＋b＋c） 2
I do not know！
A formula for finding the root of a quadratic equation with two variables
Formula method of univariate quadratic equation: change the univariate quadratic equation into a general form, and then calculate the value of the discriminant △ = b2-4ac. When b2-4ac ≥ 0, substitute the values of various coefficients a, B, C into the root formula x = [- B ± (b ^ 2-4ac) ^ (1 / 2)] / (2a), (b ^ 2-4ac ≥ 0) to get the root of the equation
If a + B = 3, ab = 2, then. 1 / 2A square + 1 / 2B square=
If a + B = 3, ab = 2, then (1 / 2a) square + (1 / 2b) square=
1 / 2A square + 1 / 2B square = 1 / 2 (A & # 178; + B & # 178;) = 1 / 2 [(a + b) & # 178; - 2Ab]
You can put in the numbers you know,
Original formula = 1 / 2 [3 & # 178; - 2 × 2] = 5 / 2
Subtract 2Ab from the square of a + B to find the sum of a square and b square, then extract a quarter, and the answer will come out,
a^4-13ab+b^4
It took me half an hour to study the brain cells
It seems very complicated, a little difficult.
It seems that I don't have the ability to
=（a²+b²）² -13ab
=（a²+b²）（a+b）（a-b）-13ab
I don't know if you can't understand it
It is known that the | a | - 1 power of equation (A-2) x = 1 is about the univariate linear equation of thought, and the value of a is obtained
It is known that the | a | - 1 power of equation (A-2) x = 1 is a linear equation of one variable with respect to the idea, and a =? x=？
The | a | - 1 power of (A-2) x = 1, so | a | - 1 = 1, so a = 2 or - 2, (A-2) x = 1, if a = 2,2-2, then 0, so determine a = - 2, and then bring it into the equation:
(-2-2)x=1
-4x=1
x= -1/4
a= -2 ,x= -1/4
once
The absolute value of a - 1 = 1
A = 2 or - 2
A=2
It's not a dollar
So a = - 2
a=-2
If a = 2, then A-2 = 0, it's not a linear equation with one variable. X can't be calculated. The equation is incomplete
a=-2 x=-1/4
A-2) x | a | - 1 power = 1, so | a | - 1 = 1, so a = 2 or - 2, (A-2) x = 1, if a = 2, 2-2 will get 0, so determine a = - 2, and then bring it into the equation:
(-2-2)x=1
-4x=1
x= -1/4
a= -2 ，x= -1/4
Define a * b = a-2b of 2a, then 3 * 4 * half=
Come on, there's no missing. The question is absolutely right
Why don't you try this,
Let 3 = a, 4 = B, bring in the definition formula and get the result 1,
Then, let the result 1 = a, half = B, and substitute it into the definition to get the result 2
Result 2 should be the final answer,
I calculated 11 / 10
Factoring high content mathematical problems with three methods factoring factor A ^ 4 + A ^ 3 + A + 1
a^4+a^3+a+1 =a^3(a+1)+(a+1)=(a+1)(a^3+1)=(a+1)^2(a^2-a+1)a^4+a^3+a+1 =a(a^3+1)+(a^3+1)=(a+1)(a^3+1)=(a+1)^2(a^2-a+1)a^4+a^3+a+1 =(a^4-2a^2+1)+(a3+2a^2+a)=(a^2-1)^2+a(a+1)^2=(a+1)^2[(a-1)^2+a]=(a+1)^2(...
It is known that the | a | - 1 power + 6 = 0 of equation (A-2) x is a linear equation of one variable with respect to x, then a = ()
∵ the | a | - 1 power + 6 = 0 of equation (A-2) x is a linear equation of one variable with respect to X
∴a-2≠0,|a|-1=1
∴a≠2,a=±2
∴a=-2
The equation of one variable is so | a | - 1 = 1, then a = ± 2
(A-2) is the X coefficient, so it cannot be equal to 0, that is, a ≠ 2
So to sum up, a = - 2
Define a * b = a-2a / 2B, then 3 * 4 * 1 / 2
It's (a-2b) / 2B
3*4=(3-8)/8=-5/8
So the original formula = - 5 / 8 * 1 / 2
=（-5/8-1)/1
=-13/5
A-2a / 2b is not clear here
3*4=3-3/4=9/4;
9/4*0.5=9/4-9/4*2=(-9/4)=-2.25
Factorization:
1.(a-2)(a^2+2a+4)
2. What is the minimum value of the algebraic formula x ^ 2-4x + 10?
3. Given that a and B are the lengths of two sides of an isosceles triangle and satisfy a ^ 2 + B ^ 2-4a-6b + 13 = 0, find the perimeter of the isosceles triangle
4. If the three sides of △ ABC are a, B, C, and satisfy a ^ 2 + B ^ 2 + C ^ 2 = AB + BC + Ca, try to judge the shape of △ ABC
5. Given x (x-1) - (x ^ 2-y) = - 2, find the value of x ^ 2 + y ^ 2 / 2-xy