Given (x + 2) &# 178; + 3y-2 = 0, find the value of 2 / 2 (x-3 / 1y & # 178;) + (- 2 / 3x + 3 / 1y & # 178;)

Given (x + 2) &# 178; + 3y-2 = 0, find the value of 2 / 2 (x-3 / 1y & # 178;) + (- 2 / 3x + 3 / 1y & # 178;)

X = - 2, y = two thirds
The first reduction is to evaluate: 2 / 2x-2 (X-2 / 1y & # 178; + 1) - (2 / 3x-3 / 1y & # 178;), where x = - 2, y = 3 / 2
2 / 2 (x-3 / 1y & # 178;) + (3 / 1y & # 178; - 2 / 3x)
=1/2x-2x+2/3y²+1/3y²-3/2x
=-3x+y²
=-3x2/3+(-2)²
=-2+4
=2
Let f (x) = 2x-3x + 1, G (x-1) = f (x), find g (x) and f {g (2)}
F (x) = 2x-3x + 1,
g（x-1）=f（x）=2x²-3x+1
Let X-1 = t
x=t+1
g(t)=2(t+1)²-3(t+1)+1=2t²+t
therefore
g(x)=2x²+x
g(2)=2×2²+2=10
f(g(2))=f(10)=2×10²-3×10+1=171
English
G (x-1) = f (x) = 2x-3x + 1
Let X-1 = t, so x = 1 + t
So g (T) = 2 (1 + T) ^ 2-3 (1 + T) + 1 = 2 + 4T + 2T ^ 2-3-3t + 1 = 2T ^ 2 + T
So g (x) = 2x ^ 2 + X
So g (2) = 10
So f [g (2)] = f (10) = 2 * 100-3 * 10 + 1 = 200-29 = 171
f(x)=2x^2-3x+1=g(x-1)
g(x)=g(x-1+1)=f(x+1)=2(x+1)^2-3(x+1)+1
=2x^2+x
g(2)=2*2^2+2=10
f[g(2)]=f(10)=2*10^2-3*10-1=169
G (x) = 2x + X;
f{g(2)}=171
Let g (x-1) = f (x), G (x-1) = 2x-3x + 1, let t = X-1, then x = t + 1, so g (T) = 2 (T + 1) - 3 (T + 1) + 1 = 2T + T, that is g (x) = 2x + X
f｛g（2）｝=f(2*2^2+2)=f(10)=171
Let t = X-1, then G (T) = f (T + 1) = 2 (T + 1) ^ 2-3 (T + 1) + 1,
That is g (x) = f (x + 1) = 2 (x + 1) ^ 2-3 (x + 1) + 1
So, G (2) = f (3) = 2 * 9-3 * 3 + 1 = 10
f{g(2)}=f(10)=2*100-3*10+1=171
Because g (x-1) = f (x), G (x-1) = 2x-3x + 1 = (x-1) (2x-1) = (x-1) [2 (x-1)]
That is g (x-1) = (x-1) [2 (x-1)]
∴g(x)=x(2x+1)
=2X + X
g(2)=10
Then f [g (2)] = f (10) = 171
Let x = x 1, then G (x) = f (x 1) = 2 (x 1) ^ 2-3 (x 1) 1 = 2x ^ 2 4x 4-3x-3 1 = 2x ^ 2 x 2
g(2)=12 f(12)=288-36 1=253
Factorization of 4-2a ^ 2B + 8ab-a
4-2a^2b+8ab-a
=8ab-2a^2b+4-a
=2ab(4-a)+(4-a)
=(2ab+1)(4-a)
（1）1/2a^2+2ab+2b^2 =(a +4ab+4b )/2 =(a+2b) /2 (2)(a+b)^2-4ab =a -2ab+b =(a-b) (3)a^2(a^2-8b^2)+16b^4 =
Factorization of X & sup2; (a + b) - a-b
x^2(a+b)-a-b
=x^2(a+b) - (a+b)
=(a+b)(x^2-1)
=(a+b)(x+1)(x-1)
（x-1）（x+1）（a+b)
x^2(a+b)-a-b=x^2(a+b)-(a+b)=(a+b)(x^2-1)=(a+b)(x+1)(x-1)
(1) 8A ^ 3-B ^ 3 (2) 20A ^ 4-33a ^ 2B ^ 2 + 7 ^ 4 factorization
1.8 (a) ^ 3-B ^ 3 = (2a) ^ 3-B ^ 3 = (2a-b) (4a ^ 2 + 2Ab + B ^ 2) 2. The original formula = 20A quartic - 5ab + 7b quartic - 28ab = 5 (4a quartic - AB) + 7 (b quartic - 4AB) = 5A (4a-b) + 7b (b-4a) = 5A (4a-b) - 7b (4a-b) = (4a-b) (5a-7b) is the partial solution of real number range
x4－2y4－2x3y＋xy3
Why not?
Original formula = X3 (x-2y) - Y3 (2y-x)
=x3(x-2y)+y3(x-2y)
=(x-2y)(x3+y3)
In addition: what is the formula for x3-y3?
Original formula = X3 (x-2y) - Y3 (2y-x)
=x3(x-2y)+y3(x-2y)
=(x-2y)(x3+y3)
=(x-2y)(x+y)(x2-xy+y2)
x3-y3=(x-y)(x2+xy+y2)
x3+y3=(x+y)(x2-xy+y2)
x*3-y*3＝（x-y）（x^2+xy+y^2）
-The third power of 2A + the square of 8a, the square of B - the square of 8ab
-2a³+8a²b-8ab²
=-2a(a²-4ab+4b²)
=-2a(a-2b)²
It's hard to understand. I have to ask you
In addition, if there is anyone who is especially good at this topic, by the way, explain the ideas and so on
The square of 1.3A + bc-3ac-ab
2. A cube-a square-2a
3.4m square - 9N square - 4m + 1
4,9-x square + 2xy-y square
(2) Factorization in real numbers
1,2x square-3x-1
2, - 2x square + 5xy + 2Y square
(3) Decompose the following factors
The square of 1,10a (X-Y) - 5B (Y-X)
2. N + 1 power of a - n power of 4A + N + 1 power of 4A
3. X cube (2x-y) - 2x + y
4.x（6x-1)-1
5.2ax-10ay+5b+6x
6.1-a squared - ab-4 / 1b squared
7. The fourth power of a + 4
You have too many questions. I can't finish them for a while. I use my mobile phone, so write the first four
1.(3a-b)(a-c)
2.a(a-2)(a+1)
3.4m(m-1)-(3n+1)(3n-1)
4.9(3+x-y)(3-x+y)
The way to solve this problem is to combine the square difference formula or the square sum formula as much as possible and simplify it to the point where it can no longer be decomposed
1,（3a-b)(a-c)
2,a(a-2)(a+1)
3. Are you wrong? I can't divide it. There's only one term with n.
oh dear..
That's too much,
This is very simple
You find the same letter in the item,
Extracted, two items together, two items together,
A new and identical part will appear,
Continue to extract,
Until there are only a few factors left in the product form. ... unfold
1,（3a-b)(a-c)
2,a(a-2)(a+1)
3. Are you wrong? I can't divide it. There's only one term with n.
oh dear..
That's too much,
This is very simple
You find the same letter in the item,
Extracted, two items together, two items together,
A new and identical part will appear,
Continue to extract,
Until there are only a few factors left in the product form. Put it away
Given the set a = {a, a + B, a + 2B}. B = {a, ax, the quadratic power of ax}, if a = B, find the value of X
Classification discussion: (1) a + B = ax, a + 2B = ax & # 178; b = ax-a substituting a + 2B = ax & # 178; a + 2 (ax-a) = ax & # 178; finishing, a (x-1) &# 178; = 0x = 1 (2) a + B = ax & # 178; a + 2B = AXB = ax & # 178; - a substituting a + 2B = AXA + 2 (AX & # 178; - a) = ax finishing, a (2x & # 178; - x-1) = 0A (x-1) (2x + 1)