It is known that the | a | - 1 power + 12 = 0 of equation (A-2) x is a linear equation of one variable with respect to x, then the solution of the equation is simple

It is known that the | a | - 1 power + 12 = 0 of equation (A-2) x is a linear equation of one variable with respect to x, then the solution of the equation is simple

If the | a | - 1 power + 12 = 0 of equation (A-2) x is a unary linear equation about X, then | a | - 1 = 1, that is | a | = 2, and A-2 ≠ 0, so a = - 2
The equation is: - 4x + 12 = 0
X=3
Definition & nbsp; if a ⁃ B = a + 2b2b, then 3 ⁃ 4 ⁃ 12=______ .
According to the meaning of the question: 3 ⁃ 4 ⁃ 12, = 3 + 2 × 42 × 4 ⁃ 12 = 118 ⁃ 12, = 118 + 2 × 122 × 12, = 118 + 1, = 238; so the answer is: 238
Solving 20 factorization problems in the second year of junior high school with answers
1．m2(p－q)－p＋q；
2．a(ab＋bc＋ac)－abc；
3．x4－2y4－2x3y＋xy3；
4．abc(a2＋b2＋c2)－a3bc＋2ab2c2；
5．a2(b－c)＋b2(c－a)＋c2(a－b)；
6．(x2－2x)2＋2x(x－2)＋1；
7．(x－y)2＋12(y－x)z＋36z2；
8．x2－4ax＋8ab－4b2；
9．(ax＋by)2＋(ay－bx)2＋2(ax＋by)(ay－bx)；
10．(1－a2)(1－b2)－(a2－1)2(b2－1)2；
11．(x＋1)2－9(x－1)2；
12．4a2b2－(a2＋b2－c2)2；
13．ab2－ac2＋4ac－4a；
14．x3n＋y3n；
15．(x＋y)3＋125；
16．(3m－2n)3＋(3m＋2n)3；
17．x6(x2－y2)＋y6(y2－x2)；
18．8(x＋y)3＋1；
19．(a＋b＋c)3－a3－b3－c3；
answer
1．(p－q)(m－1)(m＋1)．
8．(x－2b)(x－4a＋2b)．
11．4(2x－1)(2－x)．
20．(x＋3y)(x＋y)．
21．(x－6)(x＋24)．
If the equation AX2 + 2x3-b = 1 is a univariate linear equation about X, find the value of a + ab-2b2
There is no X & # 178 for the equation of degree one variable;
So a = 0
3-b=1
B=2
So the original formula = 0 + 0-8 = - 8
Given that a is not equal to B, the square of a-2a-1 = 0, the square of b-2b-1 = 0, find the value of a + B
If the square of b-2b-1 = 0 and the square of a-2a-1 = 0, find the value of a / B + B / A
Suppose a and B are two solutions of X & # 178; - 2x-1 = 0, then a + B = 2, ab = - 1, then a / B + B / a = (A & # 178; + B & # 178;) / AB = [(a + b) &# 178; - 2Ab] / AB = - 6
Factorization of (X-Y) 3 + (Y-Z) 3 + (z-x) 3
(x-y)^3+(y-z)^3+(z-x)^3
=[(x-y)^3+(y-z)^3]+(z-x)^3
=(x-y+y-z)[(x-y)^2-(x-y)(y-z)+(y-z)^2]+(z-x)^3
=(x-z)(3y^2-3xy+3xz-3yz)
=3(x-z)(x-y)(z-y)
Original formula = x ^ 3 - 3x ^ 2 * y + 3x * y ^ 2 - y ^ 3 + y ^ 3 - 3Y ^ 2 * Z + 3Y * Z ^ 2 - Z ^ 3
+ Z^3 - 3Z^2 *X + 3Z*X^2 - X^3
= 3X*Y^2 - 3X^2 *Y + 3Y*Z^2 - 3Y^2 *Z + 3Z*X^2 - 3Z^2 *X
Write a linear equation of one variable satisfying the following conditions: the coefficient of the unknown is one third, the solution of the equation is 2, and the equation of such an equation can be written as
1/3x+1/3=1
1/3(x+1)=1
x+1=3
x=3-1
X=2
1/3x-2/3=0
1/3x=2/3
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1 / 3x = C (C is the required number)
Substituting x = 2 gives C = 2 / 3
So the equation is 1 / 3x = 2 / 3
1/3x+2=8/3
Given that x = 1 is a solution of quadratic equation AX2 + bx-40 = 0, and a is not equal to B, find the value of (the square of a plus the square of B) divided by (2a-2b)
Substituting 2A + B = 40
a=(40-b）/2
Bring in the original (the latter) and do it yourself
Factorization (x + y) * 3 + (z-x) * 3 - (y + Z) * 3
(x+y)*3+(z-x)*3-(y+z)*3
=(x+y)*3+[(z-x)-(y+z)][(z-x)^2+(z-x)(y+z)+(y+z)^2]
=(x+y)*3+[-(x+y)](z^2-2xz+x^2+yz+z^2-xy-xz+y^2+2yz+z^2)
=(x+y)*3-(x+y)(3z^2+x^2+y^2-3xz+3yz-xy)
=(x+y)[(x+y)^2-(3z^2+x^2+y^2-3xz+3yz-xy)]
=(x+y)(x^2+2xy+y^2-3z^2-x^2-y^2+3xz-3yz+xy)
=(x+y)(3xy+3xz-3yz-3z^2)
=3(x+y)[x(y+z)-z(y+z)]
=3(x+y)(y+z)(x-z).
(x+y)*3+(z-x)*3-(y+z)*3
You are using * to express power. Here we usually use ^ to express the power and * to express the multiplication. Please do as the Romans do.
(x+y)^3 + (z-x)^3 - (y+z)^3 =
x^3 + 3x^2y + 3xy^2 + y^3 + z^3 -3z^2x + 3zx^2 - x^3 - y^3 - 3y^2z -3yz^2 - z^3
Write a linear equation with one variable, so that its solution is minus 3 / 2, and the coefficient of the unknown is a positive integer, then the equation is------------------
2X + 3 = 0, or a multiple of it
2x+3=0
2x+3=0
2x-1=-4