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Let x ^ 2 + 4x + y ^ 2-6y + 13 = 0. Find the value of (x-2y) divided by (x ^ 2 + y ^ 2)
X ^ 2 + 4x + y ^ 2-6y + 13 = 0 can be converted into
(X+2)^2+(y-3)^2=0
So x = - 2
Y=3
(x-2y) divided by (x ^ 2 + y ^ 2) = - 8 / 13
X ^ 2Y ^ 2-4x-6y + 13 = 0 find the value of (X-Y) ^ - 2 Reward can be given more accurately
Your question should be: x ^ 2 + y ^ 2-4x-6y + 13 = 0, find the value of (X-Y) ^ - 2
x^2+y^2-4x-6y+13=0
（x^2-4x+4）+（ y^2-6y+9）=0
(x-2)^2+(y-3)^2=0
x-2=0,y-3=0
x=2,y=3
(x-y)^-2=1/（x-y）^2=1/(2-3)^2=1
If x ^ 2-4x + y ^ 2-6y + 13 = 0, find the value of x ^ 2Y
y=3 X=2.
Given x * 2-4x + y * 2 + 6y + 13 = 0, find the value of X * 2 + 2Y It's urgent. Thank you
(x^2-4x+4)+(y^2+6y+9)=0
(x-2)^2+(y+3)^2=0
So x = 2, y = - 3
17.3x+8y=51 x+6y=27 18.9x+3y=99 4x+7y=95 19.9x+2y=38 3x+6y=18 20.5x+5y=45 7x+9y=69 21.8x+2y=28 7x+8y
17. (2) * 3 - (1) = 3x + 18y-3x-8y = 30, get 10Y = 30, get y = 3, bring y = 3 into 3x + 24 = 51 in (1), get x = 9.18. (1) * 4 - (2) * 9 = 36x + 12y-36x-63y = 99 * 4-95 * 9, arrange - 51y = - 459, get y = 9, put y = 9 into 9x + 3 * 9 = 99 in (1), get x = 8.19. (1)
Solve the equation 2 / 3x + 1 / 2Y = - 5 / 24 1 / 4x-1 / 6y = - 1 / 6 If you don't know the writing clearly, ask me ~! Speed! Process! Process! The first equation and the second equation are connected with braces {2 / 3x + 1 / 2Y = - 5 / 24 ① {1/4x-1/6y=-1/6②
Multiply the two sides of the first equation by 24 to get 16x + 12Y = - 5 ------ (3)
Multiply the two sides of the second equation by 12 to get 3x-2y = - 2 ------ (4)
（2）*6,18x-12y=-12------（5）
（3）+（5）,34x=-17,x=-1/2；
X = - 1 / 2 into (3), - 8 + 12Y = - 5, y = 1 / 4
Given that 4x ^ 2 + y ^ 2-4x + 6y + 10 = 0, find the value of (3x-2y) ^ 3
4x^2+y^2-4x+6y+10=0
4x^2-4x+1+y^2+6y+9=0
The formula is as follows
(2x-1)²+(y+3)²=0；
So:
2x-1=0
y+3=0
Namely:
x=1/2
y=-3
The result is as follows:
(3x-2y)^3
=（3/2+6）^3
=（15/2）^3
=3375/8
Given (x + 2) + | y + 1| = 0, find the value of 3x - [3x squared Y - (4x-2y)] + 4x's squared y + 6y
X + 2 = 0, x = - 2, | y + 1, | 0, y = - 1, substituting 3x - [3x's Square Y - (4x-2y)] + 4x's Square y + 6y
3 * (- 2) - [3 * (- 2) ^ 2 * (- 1) - 4 * (- 2) + 2 * (- 1)] + 4 * (- 2) ^ 2 * (- 1) + 6 * (- 1)
=-22
Let n be a natural number and x ^ 2-4x + y ^ 2-6y + 13 = 0. Find (3x-2y) ^ 2n + 1 As the title
∵x^2-4x+y^2-6y+13=0
∴(x-2)^2+(y-3)^2=0
∴x=2,y=3
(3x-2y)^2n+1=0^2n+1=1
Given the square of (3x-2y + 1) + 4x-3y-3 │ = 0, find x, y
The absolute sum of squares is greater than or equal to 0
Add up to 0
If one is greater than 0, then the other is less than 0, impossible
So both are equal to zero
So 3x-2y + 1 = 0 (1)
4x-3y-3=0 (2)
(1)*4-(2)*3
-8y+9y+4+9=0
y=-13
x=(2y-1)/3=-9
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