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If the real numbers x and y satisfy 3x ^ 2 + 6 and Y ^ 2 = 6x, the value ranges of X and x ^ 2 + y ^ 2 are obtained respectively
y²=-x²/2+x
y²>=0
Then - X & sup2 / / 2 + x > = 0
x²-2x=x(x-2)
If real numbers x and y satisfy 3x * 2 + 2Y * 2 = 6x, then the maximum direct value of X * 2 + y * 2 is?
2y^2=-3x^2+6x
2y^2>=0
So - 3x ^ 2 + 6x > = 0
x^2-2x
Given X & # 178; + Y & # 178; + 6x-2y + 10 = 0, find - 4 / 3x + y
x^2+6x+9+y^2-2y+1=0
(x+3)^2+(y-1)^2=0
x+3=0
y-1=0
So x = - 3, y = 1
therefore
-4/3x+y=-4/(-9+1)=1/2
{3x+2y=8,6x-5y=-2 y
3x+2y=8(1),6X-5y=-2(2)
(1) Multiply it by 2 to get 6x + 4x = 16 (3)
Subtract (2) from (3) to get 6x + 4Y - (6x-5x) = 16 - (- 2)
y=2
Substituting y = 2 into (1), we get x = 4 / 3
So x = 4 / 3, y = 2
If 3x ^ + 2Y ^ = 6x, find the maximum value of x ^ + y ^
^It's better to have a process
y^=(6x-3x^)\2
x^+y^=-x^\2+3x=-(x-3)^\2+9\2
The maximum value is 9-2, and there is no minimum value
Factorization: 3x ^ 2 + 6xy ^ 2-12xy
yms x y
>> factor(3*x^2+6*x*y^2-12*x*y)
ans =
3*x*(x+2*y^2-4*y)
Factorization - 2x + 1 / 2Y
The original formula = (1 / 2) * 2 (- 2x ^ 2 + 1 / 2Y ^ 2)
=(1/2)(y^2-4x^2)
=(1/2)(y-2x)(y+2x)
=(1/2y-x)(y+2x)
Factorization of 2x ^ 2 + 4xy + 2Y ^ 2-8z ^ 2
Original formula = 2 (x + y) ^ 2-8z ^ 2 = 2 [(x + y) ^ 2 - (2Z) ^ 2] = 2 (x + y + 2Z) (x + y-2z)
Factorization x ^ 2-2x-2y ^ 2 + 1
= x^2-2x+1-2y^2
=(x-1)^2-2y^2
=(x-1+√2y)(x-1-√2y)
Factorization-6x ^ 2 + 4xy-2x
[extract common factor - 2x]
-6x^2+4xy-2x
=-2x(3x-2y+1)
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