√ (x-y)^3 (y-z)^3 (z-x)^3=3(x-y)(y-z)(z-x) 220861
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Complete cubic parametrization of the Fermat cubic surface w 3 x 3 y 3 z 3 = 0 This is a famous Diophantine problem, to which Dickson's History of the Theory of Numbers, Vol II devotes many pages It is usually phrased as w 3 x 3 y 3 =z 3 or w 3 x 3 =y 3 z 3, with the implication that the variables are to be positive, as in the integer solutions 3 3 4 3 5 3 =6 3 (an amusing3 Describe geometrically the set of points (x,y,z) that satisfy y = −3 4 Describe geometrically the set of points (x,y,z) that satisfy x y = 2 5 The equation x y z = 1 describes some collection of points in R3 Describe and sketch the points that satisfy x y z = 1 and are in the xy plane, in the xz plane, and in the yz plane 6
(x-y)^3 (y-z)^3 (z-x)^3=3(x-y)(y-z)(z-x)
(x-y)^3 (y-z)^3 (z-x)^3=3(x-y)(y-z)(z-x)-Subtracting (A){(B) we get ‚(x ¡ 2y) = 0, so either ‚ = 0 or x = 2yBut ‚ = 0 would give x = y = z = 0, and f(0;0;0) = 0 is obviously not the maximumTherefore we work with x = 2y Subtracting (B){ we get ‚(2y¡3z) = 0, and since we already discarded the case ‚ = 0 we are left with z = 2 3 y Using the results in the two frames into (D) we getZ 1 0 (x2 − 4x 3)dx = 4/3 (b) F(x,y,z) = xiy j(x2 y2)k, C is the boundary of the part of the paraboloid z = 1 − x 2− y in the first octant Solution The curl of F is curlF = ∂ i j k ∂x ∂ ∂y ∂z x y x 2 y = 2y i − 2xj The surface S can be represented as r
Prove That 2x 3 2y 3 2z 3 6xyz X Y Z X Y 2 Y Z 2 Z X 2 Brainly In
Solución x = 3, y = 2, z = 1 Son tres planos que se cortan en el punto (3, 2, 1) 2 a) Resuelve el sistema b) Añade una tercera ecuación de modo que siga siendo compatible c) Añade una tercera ecuación de modo que sea incompatible d) Interpreta geométricamente loIt's a simple miscalculation (8,1,1)\times(1,1,1) = (0,9,9), and normalized \frac{1}{\sqrt{2}}(0,1,1) Let x,y,z be nonnegative real numbers satisfying the condition xyz=1 The maximum possible value of x^3y^3y^3z^3z^3x^3 has the form \dfrac{a}{b}, where a and b are positive, coprime = 2x 3 2y 2 z 2xz 2 2x 2 y 2xyz 2x 2 z 2x 2 y 2y 3 2yz 2 2xy 2 2xyz 2y 2 z 2x 2 z 2y 2 z 2z 3 2yz 2 2xyz 2xz 2 = 2x 3 2y 3 2z 3 6xyz = 2 ( x 3 y 3 z 3 3xyz ) Hence LHS = RHS ( Hence proved )
(x;y) 2 D = (x;y) 2 R2 x2 (y ¡1)22 • 1 ¾ 3 Calculelaintegral Z SFor &ve variables x;y;z;t;s Two lines intersect each other if and only If this system has a solution If, for instance, (x0;y0;z0;t0;s0)is a solution, then the &rst three components, (x0;y0;z0)is a point of intersection 4 We now proceed to solution the system by eliminating x,y,zTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW 2x 3 y 3z =5 , x 2y z=4 , 3x y2z = 3
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Answer by You can put this solution on YOUR website! Best answer We have Now, we know that If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants Taking out common factor (y – x) from R2 and (z – x) from R3, we get
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