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The curve represented by equation (3x-4y-12) [log2 (x + 2Y) - 3] = 0 passes through the points a (0, - 3), B (4,2), C (4,0), D (5 / 3, - 7 / 4), where Why aren't a and D?
A（0,-3）,(x+2y)0 ,x+2y=8 ,[log2(x+2y)-3]=0 ,（3x-4Y-12)=-8
C（4,0）,x+2y>0 ,x+2y=4,[log2(x+2y)-3]=-1,（3x-4Y-12)=0
D（5/3,-7/4),x+2y
The curve represented by equation x ^ 2-y ^ 2-4x ^ 2 + 4Y ^ 2 = 0 is
It is reduced to y ^ 2 = x ^ 2
So: y = x or y = - x
Therefore, the curve represented by the original equation is two straight lines: y = x and y = - X
The line represented by equation x ^ 4-y ^ 4-4x ^ 2 + 4Y ^ 2 = 0 is?
Wrong number. Sorry
What is the curve?
Circle or something
x^4-y^4-4x^2+4y^2=x^4-4x^2＋4 －4 -y^4+4y^2－4 ＋ 4
＝（x²-2)² - (y²-2)²＝0
So there is x-178; - 2 = y-178; - 2 or x-178; - 2 = 2-y-178;
That is, X & # 178; = y & # 178; or X & # 178; + Y & # 178; = 4
That is, two straight lines y = x, y = - X and a circle with the origin as the center and radius of 2
The curve represented by equation 4x ^ 4-4y ^ 4-x ^ 2 + y ^ 2 is
A two intersecting lines and a circle
B two parallel lines and a circle
C two points and a circle
D one point and two circles
Choose a
The equation can be changed to (x ^ 2-1 / 8) ^ 2 = (y ^ 2-1 / 8) ^ 2
That is y = + - X
x^2+y^2=0
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