Let {(the square of X + the square of Y) - - the square of the difference of (X-Y) + 2Y (X-Y)} divided by 4Y = 1 Find the square of 4x / 4x - the square of Y minus 1 / 2x + y

Let {(the square of X + the square of Y) - - the square of the difference of (X-Y) + 2Y (X-Y)} divided by 4Y = 1 Find the square of 4x / 4x - the square of Y minus 1 / 2x + y

{(x2+y2)-(x-y)2+2y(x-y)}/4y=1
{ x2+y2- (x2+y2-2xy)+2xy-2y2}/4y=1
(4xy-2y2)/4y=1
The result is: 2x-y = 2
4x/( 4x2-y2)-1/(2x+y)
=4x/(2x+y)(2x-y) -1/(2x+y)
=[4x-(2x-y)]/ (2x+y)(2x-y)
=(2x+y) / (2x+y)(2x-y)
=1/(2x-y)
=1/2

① - x + 2Y X-Y divided by X squared + 4xy + 4Y squared X-Y squared ② 4-4x + 2x-6 of X divided by (X-2) (x + 3) 3-x

① - x + 2Y fraction X-Y divided by X squared + 4xy + 4Y squared x squared - y squared
= -(x-y)/(x+2y) ÷ (x^2-y^2)/(x^2+4xy+4y^2)
= -(x-y)/(x+2y) ×(x^2+4xy+4y^2)/(x^2-y^2)
= -(x-y)/(x+2y) ×(x+2y)^2/[(x+y)(x-y)]
= -(x+2y)/(x+y)
② 4-4x + X 2x-6 divided by (X-2) (x + 3) by 3-x
= (2x-6)/(4-4x+x^2) ÷ (3-x)/[(x-2)(x+3)]
= 2(x-3)/(x-2)^2 × (x-2)(x+3)/(3-x)
= 2(x-3)/(x-2)^2 × (x-2)(x+3)/[-(x-3)]
= -2(x+3)/(x-2)

Find the intersection point of the straight line l1:3x + 2y-1 = 0, l2:5x + 2Y + 1 = 0, and vertical L3: 3x-5y + 6 = 0

Simultaneous 3x + 2y-1 = 0 ①
5x+2y+1=0 ②
② - ① get the coordinates of intersection point x = - 1, y = 2
If it is perpendicular to 3x-5y + 6 = 0, then k = - 1 / (3 / 5) = - 5 / 3,
The linear equation is Y-2 = - 5 / 3 (x + 1), that is, 5x + 3y-1 = 0

The linear equation passing through the intersection point of l1:3x-5y-10 = 0 and L2: x + y + 1 = 0 and parallel to L3: x + 2y-5 = 0 is______ .

Through the intersection point of l1:3x-5y-10 = 0 and L2: x + y + 1 = 0, the slope of the straight line x = 58, y = - 138 and L3: x + 2y-5 = 0 is: − 12. The linear equation: y + 138 = - 12 (x − 58) is: 8 x + 16y + 21 = 0

The equation of the straight line passing through the intersection of the straight lines 3x + 2Y + 6 = 0 and 2x + 5y-7 = 0 and with equal intercept on the two coordinate axes is obtained

When the straight line L passes through (- 4,3) and passes through the origin, because the intercept of the line L on the two coordinate axes is equal, let y = KX, replace (- 4,3) to obtain k = - 34

The plane equations passing through the point P (2,0, - 3) and perpendicular to the straight lines x-2y + 4z-7 = 0 and 3x + 5y-2z + 1 = 0 are solved

Two normal vectors:
(1,-2,4),(3,5,-2)
The surface normal vector is the cross product of the two
n=(1,-2,4),(3,5,-2)=(4-20,12+2,5+6)=(-16,14,11)
The plane equation obtained is
-16(x-2)+14y+11(z+3)=0
Namely
-16x+14y+11z+65=0
This is the most direct way I can think of

The equation of the line L passing through the intersection of the lines l1:3x + 2y-1 = 0 and l2:5x + 2Y + 1 = 0 and perpendicular to the line l3:3x-5y + 6 = 0 is obtained

Simultaneous equations
3x+2y-1=0
5x+2y+1=0 ，
The solution
x=-1
Y=2
The coordinates of intersection points of L1 and L2 are (- 1, 2),
From the slope 3 of L3
The slope of L is - 5
3，
The equation of the straight line is Y-2 = - 5
3（x+1），
By changing it into a general formula, we can get 5x + 3y-1 = 0

Given that 3x-2y-5z = 0,2x-5y + 4Z = 0, and X, y, Z are not 0, find the value of 3x * x + 2Y * y + 5Z * Z / 5x * x + y * y-9z * Z

[explanation]
Consider Z as a constant. From the known two equations, we can get the solution
x=3z
y=2z
Substituting it into the expression to be evaluated, the
3x*x+2y*y+5z*z/5x*x+y*y-9z*z
=[3(3z)^2+2(2z)^2+5z^2]/[5(3z)^2+(2z)^2-9z^2]
=40z^2/40z^2
=1

To solve the system of three variable linear equations {3x-5y + 6Z = 4, 3x-2y + 2Z = 3, - 3x-3y + 5Z = - 1

3x-5y+6z=4 (1)
3x-2y+2z=3 (2)
-3x-3y+5z=-1 (3)
(1)-(2)
-3y+4z = 1 (4)
(2)+(3)
-5y+7z=2 (5)
5(4)-3(5)
z = 1
from (4) y=1
from (1)
3x-5+6=4
x= 1
IE
x=y=z=1

The substitution method is used to solve the binary system of first order equations: 1, 2x + y = 32, 2x + 7Y + - 13, 2x-y = 6, 3x-5y = 11, x-2y = 5x + 2Y + - 24, 2x = 3Y, 3Y = x + 2

The second equation is wrong, isn't it "2x + 7Y = - 1"?