What can (x + 2Y) &# 178; (x-2y) &# 178; - (2x + y) &# 178; (2x-y) &# 178; be reduced to?

What can (x + 2Y) &# 178; (x-2y) &# 178; - (2x + y) &# 178; (2x-y) &# 178; be reduced to?

(x+2y)²(x-2y)²-(2x+y)²(2x-y)²
=[(x+2y)(x-2y)]²-[(2x+y)(2x-y)]²
=(x^2-4y^2)^2 - (4x^2-y^2)^2
=(x^2-4y^2 + 4x^2-y^2) * ( x^2-4y^2 - 4x^2+y^2 )
=(5x^2-5y^2)*(-3x^2-3y^2)
=5(x^2-y^2)*(-3)(x^2+y^2)
=-15(x+y)(x-y)(x^2+y^2)

|x-2y+1|+(2x-y-5)²=0

According to the absolute value and square are non negative
X-2y + 1 = 0 means x-2y = - 1
2x-y-5 = 0, that is, 2x-y = 5
① * 2 - 2
-4y+y=-2-5
3y=7
y=7/3
It's good for people
x=2y-1=14/3-1=11/3
therefore
x=11/3
y=7/3

2X & sup2; + Y & sup2; + 2xy-2y + 2 = 0 find x + y = ()

Wrong, it's 2x & sup2; + Y & sup2; + 2xy-2x + 1 = 0
That is, (X & sup2; + 2XY + Y & sup2;) + (X & sup2; - 2x + 1) = 0
(x+y)²+(x-1)²=1
So x + y = 0, X-1 = 0
So x + y = 0

The {3x-2t = 5Y + 1.2x-1 = 4y-3t} denotes y with an algebraic expression containing X,

2t=3x-5y-1;
t=(3x-5y-1)/2;
3t=4y-2x+1;
t=(4y-2x+1)/3;
3(3x-5y-1)=2(4y-2x+1);
9x-15y-3=8y-4x+2;
13x-23y-5=0;
y=(13x-5)/23;

Factorization of 6x ^ 2 + 7xy + 2Y ^ 2-8x-5y + 2
Such as the title

6X ^ 2 + 7xy + 2Y ^ 2-8x-5y + 2 = (2x + y) (3x + 2Y) - 8x-5y + 2 next, we use the undetermined coefficient method to solve (2x + y) (3x + 2Y) - 8x-5y + 2 = (2x + y + m) (3x + 2Y + n) = 6x ^ 2 + 7xy + 2Y ^ 2 + (3m + 2n) x + (n + 2m) y + mnmn = 23m + 2n = - 8N + 2m = - 5N = - 1m = - 2 (2x + y) (3x + 2Y) - 8x-5y + 2 = (2x + Y-2)

Decomposition factor: (x + 2Y) (x-2y) - (x-4y) (x + 4Y) + (6y-5x) (6x-5y)

(x+2y)(x-2y)-(x-4y)(x+4y)+(6y-5x)(6x-5y)
= x^2-4y^2-x^2+16y^2+36xy-30y^2-30x^2+25xy
= -18y^2+61xy-30x^2

All integer solutions of 11x + 5Y = 7
Another question: 4x + y = 3xy
500 points for correct answer! I'll share more

11x+5y=7
There are countless integer solutions!
x=-3,y=8
x=-13,y=30
x=-23,y=52
x=33,y=74
x=43,y=96
.
x=7,y=-14
x=17,y=-36
x=27,y=-58
x=37,y=-80
x=47,y=-102
.
4x+y=3xy
x=0,y=0
x=1,y=2
x=-1,y=1

3x-5y+3z=0 3x+7y-6z=0

Reference: given {x-5y + 2Z = 0, 3x-5y-4z = 0, find X: Y: Z
(2) - (1) get:
2x-6z=0；
x=3z；
Bring in (1)
5y=5z；
y=z；
So x: Y: z = 3:1:1;

When a equals minus one, the cube of 4A equals zero

(a^2-9)/(a^2-4a+3)×（a-1)/a^2+4a+3
(2a^2-3a+1)/（1-a^2)÷(4a^2-4a+1)/(2a^2+a-1)
(3x+9)/(4-4x+x^2)÷（x+3）/(x-2)

(a^2-9)/(a^2-4a+3)×（a-1)/a^2+4a+3
=[(a-3)(a+3)/(a-3)(a-1)×(a-1)/[(a+3)(a+1)]
=(a+3)/(a-1)×(a-1)/[(a+3)(a+1)]
=1/(a+1)