√70以上 x^2 y^2 z^2/4 2xy-yz-zx 251295-(1)/(4)x^(2)+y^(2)+(1)/(16)x^(2)-xy-(1)/(2)y+(1)/(4)x
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(1)/(4)x^(2)+y^(2)+(1)/(16)x^(2)-xy-(1)/(2)y+(1)/(4)x-The idea is to complete the square x2 y2 z2 xyxz yz = (x 21 (yz))2 y2 z2 − 41 (y z)2 yz = (x 21 (y z))2 43 y2 43 z2 21 yz = (x 21 (y z))2 43 (y 31 z)2 32 z2 Simultaneous diagonlisation of two quadratic forms, one of which is positive definite The common way to solve this kind of ODE is change of function z = y u ( y) y d u d y u = 1 − u 1 u This is a separable ODE Then, a few calculus leads to z 2 2 y z − y 2 = c 2 The general solution of the PDE can be implicitly expressed on various equivalent forms, for example Φ ( ( x 2 y 2 z 2), ( z 2 2 y z − y 2)) = 0
(1)/(4)x^(2)+y^(2)+(1)/(16)x^(2)-xy-(1)/(2)y+(1)/(4)xのギャラリー
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「(1)/(4)x^(2)+y^(2)+(1)/(16)x^(2)-xy-(1)/(2)y+(1)/(4)x」の画像ギャラリー、詳細は各画像をクリックしてください。
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「(1)/(4)x^(2)+y^(2)+(1)/(16)x^(2)-xy-(1)/(2)y+(1)/(4)x」の画像ギャラリー、詳細は各画像をクリックしてください。
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Dự đoán khi \(x=y=z=\frac{1}{\sqrt{3}}\) thì ta tìm được \(P=\frac{3\sqrt{3}}{2}\) Ta sẽ chứng minh nó là GTNN Thật vậy, ta cần chứng minhWe have x2 y2 = 36−z2 and x y = 10−z, which gives (10−z)2 −2xy = 36−z2 or xy = 32− 10z z2 and xyz = 32z − 10z2 z3 Also, (x y)2 ≥ 4xy, which gives 3z2 −z 28 ≤ 0 or 2 ≤ z ≤ 314 Find the equation of cone whose generators pass through the section of the sphere and plane?
Incoming Term: x^2+y^2+z^2/4+2xy-yz-zx, factorize x^2+y^2+z^2/4+2xy-yz-zx, (1)/(4)x^(2)+y^(2)+(1)/(16)x^(2)-xy-(1)/(2)y+(1)/(4)x,
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