-Using square difference formula to decompose factor ①-4x²+（2x-3y）² ②16（a+b）²-9（a-b）² ③9m²（a-b）^3+49（b-a）^3 ④（x+y-z）²-（x-y+z）² ⑤（a²-b²）+（3a+3b） ⑥x^4-1

-Using square difference formula to decompose factor ①-4x²+（2x-3y）² ②16（a+b）²-9（a-b）² ③9m²（a-b）^3+49（b-a）^3 ④（x+y-z）²-（x-y+z）² ⑤（a²-b²）+（3a+3b） ⑥x^4-1

solution
-4x²+(2x-3y)²
=(2x-3y)²-(2x)²
=(2x-3y-2x)(2x-3y+2x)
=-3y(4x-3y)
16(a+b)²-9(a-b)²
=[4(a+b)]²-[3(a-b)]²
=[4(a+b)-3(a-b)][4(a+b)+3(a-b)]
=(a+7b)(7a+b)
9m²(a-b)³+49(b-a)³
=9m²(a-b)³-49(a-b)³
=(a-b)³[(3m)²-7²]
=(a-b)³(3m-7)(3m+7)
(x+y-z)²-(x-y+z)²
=[(x+y-z)-(x-y+z)][(x+y-z)+(x-y+z)]
=(2y-2z)(2x)
=4x(y-z)
(a²-b²)+(3a+3b)
=(a-b)(a+b)+3(a+b)
=(a+b)(a-b+3)
x^4-1
=(x²)²-1²
=(x²-1)(x²+1)
=(x-1)(x+1)(x²+1)

How to use the square difference formula to calculate (2x + 3y-5) (2x-3y + 1)?

=[(2x-2)+(3y-3)][(2x-2)-(3y-3)]
=(2x-2)(2x-2)-(3y-3)(3y-3)
=...

It is calculated by the complete square formula: (1) (4x-3y) (2) (2) (2b of 1.5a-3) (3) 63? (4) 98 Wrong question. It's (1) (4x-3y) 2 (2) (- 2m-1) 2 (3) (1.5a-2b of 3) (4) 63? (5) 98

(4x-3y) mm2 = 16x? - 24xy + 9y? 2 (1.5a-2b of 3)? = (3 / 2a-2 / 3b) 2 = 9 / 4A? - 2Ab + 4 / 9b? 63? = (60 + 3) mm2 = 3600 + 360 + 9 = 396998? = (100-2) mm2 = 10000-400 + 4 = 9604

(- 4x + 3Y) 2 complete square formula

（-4x+3y）²
=（-4x）²-2×4x×3y+（3y）²
=16x²-24xy+9y²

The second power of (4x-3y) uses the complete square formula to calculate the main process

(4x-3y)^2
=(4x)^2-24xy+(3y)^2
=16x^2-24xy+9y^2

(- 4x + 3Y) ^ 2 is calculated using the complete square formula

（-4x+3y）²
=16x²-24xy+9y²

Calculate (4x-3y) with the complete square formula By the way, tell me about the algorithm of complete square formula, formula, key points!

=4X (4x-3y) - 3Y (4x-3y) = 16x square - 12xy - (12xy-9y Square) = 16x square + 9y square

Given that x 2 + 4Y squared - 2x = 4Y-2, find the value of (2x-3y) 2 - (3y-x) 2

It is known that x 2 + 4Y squared - 2x = 4Y-2
x^2-2x+1+4y^2-4y+1=0
(x-1)^2+(2y-1)^2=0
x=1,y=1\2
（2x-3y）²-（3y-x）²
=(2-3\2)^2-(3\2-1)^2
=0

Given x 2 + 4Y + y 2 - 6y + 13 = 0, find the value of 2x + 3Y

It is known that x 2 + 4x + y 2 - 6y + 13 = 0
Then (x + 2) 2 + (Y-3) 2 = 0
So x + 2 = 0, Y-3 = 0
So x = - 2, y = 3
So 2x + 3Y = 2 * (- 2) + 3 * 3 = 5
If you don't understand, I wish you a happy study!

To solve the equations: x 2 - 4Y 2 + X + 3y-1 = 0 and 2x-y-1 = 0

x^2-4y^2+x+3y-1=0 (1)
2x-y-1=0 (2)
From (2)
y=2x-1.(3)
By substituting (1), we can get
x^2-4(2x-1)^2+x+3(2x-1)-1=0.
Namely
-15x^2+23x-8=0.
The solution
x1=1,x2=8/15.
By substituting X1 and X2 into (3), the solutions of the original equations are obtained as follows
x1=1,
y1=1.
x2=8/15,
y2=1/15.