Calculation: (2x-3y) 2 (2x + 3Y) 2 （2x-3y)²(2x+3y)² =（4x²-12xy+9y²)(4x²+12xy+9y²) =[4x²-(12xy-9y²)] [4x²+(12xy+9y²)] =16x^4-(12xy-9y²)² Let's see if there is any problem with this method

Calculation: (2x-3y) 2 (2x + 3Y) 2 （2x-3y)²(2x+3y)² =（4x²-12xy+9y²)(4x²+12xy+9y²) =[4x²-(12xy-9y²)] [4x²+(12xy+9y²)] =16x^4-(12xy-9y²)² Let's see if there is any problem with this method

（2x-3y)²(2x+3y)²
=[(2x-3y)(2x+3y)]²
=(4x²-9y²)²
=16x⁴-72x²y²+81y⁴
Your method is no problem, but it becomes more complicated if you want to use the formula

The result of (2x-3y) 2 (2x + 3Y) 2 is calculated

（2x-3y）²（2x+3y）²
=（2x-3y）（2x+3y）*（2x-3y）（2x+3y）
=(4x²-9y²)²
=16x^4+81y^4-72x²y²

Calculation: 8 x / 3 y · (- 3 y / 2 x 2) 3

=8x / 3Y × (- 27y? / 8x to the 6th power)
=-The 5th power of 9y ^ / X

Calculate (2x-3y) 2 (2x + 3Y) 2 = (square difference)

(2x-3y)²(2x+3y)²
=[(2x-3y)(2x+3y)]²
=(4x²-9y²)²

0.5x-2=0.8(2x-1.

0.5x-2=0.8(2x-1.5)
0.5x-2=1.6x-1.2
1.6x-0.5x＝1.2-2
1.1x＝-0.8
x＝-8/11

If 2x2-5x + 8 If 2x2 − 5x + 1-5 = 0, then the value of 2x2-5x-1 is______ .

Let y = 2x2-5x + 1, then the original formula can be transformed into:
y-1+8
y-5=0
The result is: y2-y + 8-5y = 0
That is to say, 6y = 0
ν y = 2 or 4
It is proved that y = 2 or 4 is the solution of the equation
Then 2x2-5x-1 = 2x2-5x + 1-2 = Y-2
Then the value of 2x2-5x-1 is 2 or 0

When x = 6, find the value of 5x-2x + 6.8

5x-2x+6.8
=3x+6.8
=3×6+6.8
=24.8

If x ^ 2-2x-1 = 0, find the value of x ^ 4 + x ^ 3-5X ^ 2-7x + 5

x²=2x+1
Then the following x 2 is replaced by 2x + 1
x³=x(x²)
=x(2x+1)
=2x²+x
=2(2x+1)+x
=5x+2
x^4=(x²)²
=(2x+1)²
=4x²+4x+1
=4(2x+1)+4x+1
=12x+5
The original formula = (12x + 5) + (5x + 2) - 5 (2x + 1) - 7x + 5
=12x+5x+2-10x-5-7x+5
=2

(1)6x^2-7x+1=0 (2)5x^2-9x-18=0 (3)4x^2-3x=52 (4)5x^2=4-2x

(1) 6 x ^ 2-7x + 1 = 0 (6x-1) (x-1) = 0, then x = 1 or 1 / 6 (2) 5x ^ 2-9x-18 = 0 (5x + 6) * (x-3) = 0, x = - 6 / 5 or 33) 4x ^ 2-3x = 52 4x ^ 2-3x-52 = 0 (x-4) (x + 13) = 0 x = 4 or-135x ^ 2 + 2X-4 = 0 5 (x + 1 / 5) ^ 2-4-5 * (1 / 5) ^ 2 = 0 5 (x + 1 / 5) ^ 2 = 0 5 (x + 1 / 5) ^ 2 = 0 (x + 1 / 5) ^ 2 = 0 5 (x + 1 / 5) ^ 2 = 2 = 0 (x + 1 / 5) ^ 2 = 2 = 5 (x 21 / 5 (x + 1 / 5) ^ 2 = 21 / 25

Solve the following bivariate linear equations: 6x-7x + 1 = 0 and 5x-18 = 9x and 4x-3x = 52

6x^2-7x+1=0
(6x-1)(x-1)=0
x1=1/6;x2=1
5x square - 18 = 9x becomes
5x^2-9x-18=0
(5x+6)(x-3)=0
x1=-6/5;x2=3
4X ^ 2-3x = 52 becomes
4x^2-3x-52=0
(4x+13)(x-4)=0
x1=-13/4;x2=4