The second power of [(x + 3Y) (x-3y)]

The original formula = (X & sup2; - 9y & sup2;) & sup2;
=（X²）²-2×X²×9Y²+（9Y²）²
=X＾4-18X²Y²+81Y＾4

Given 2x & sup2; - 3Y & sup2; = XY, (XY ≠ 0), evaluate the value of (4xy-x & sup2;) / (XY + 2Y & sup2;)

From 2x & sup2; - 3Y & sup2; = XY, (x + y) (2x-3y) = 0
So x = - y, or 2x = 3Y,
When x = - y, we substitute = - 5
When 2x = 3Y, we can get 15 / 14

2X ^ 2 + 5xy + 3Y ^ 2 = 5, k = 2x + 3Y + XY

2x^2+5xy+3y^2=0
(x+y)(2x+3y)=0
X = - y or x = - 3Y / 2
The maximum value is 1 / 4; k = 2x + 3Y + xy = Y-Y ^ 2 = - (Y-1 / 2) & # 178; + 1 / 4;
Or, k = 2x + 3Y + xy = - 3Y & # / 2, the maximum value is 0

Given the algebraic formula 2x ^ 2 + xy = 8, y ^ 2-xy = - 2, find the value of 2x ^ 2 + y ^ 2 and 4x ^ 2-3y ^ 2 + 5xy

The solution is as follows!
2x^2+xy=8,（1）y^2-xy=-2（2）
By adding the two formulas, we can get 2x ^ 2 + y ^ 2 = 6
(1) * 2 - (2) * 3 gives 4x ^ 2-3y ^ 2 + 5xy = 22

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