a2+b2=(a+b)2-2ab,
a2+b2=(a-b)2+2ab,
(a+b)2-(a-b)2=4ab,
a2+b2+c2=(a+b+c)2-2(ab+ac+bc)
Second, the application of multiplication formula deformation
Example 1: It is known that x2+y2+4x-6y+ 13=0, and both x and y are rational numbers. Find the value of xy.
Analysis: Reverse the complete power formula and put it
X2+y2+4x-6y+ 13 is the sum of two completely flat modules, and the values of x and y can be obtained by using the nonnegativity of completely flat modules.
Solution: ∫x2+y2+4x-6y+ 13 = 0,
(x2+4x+4)+(y2-6y+9)=0,
That is, (x+2)2+(y-3)2=0.
∴x+2=0,y=3=0。
That is, x=-2 and y=3.
∴xy=(-2)3=-8。
Analysis: This question is skillfully used.
Example 3 is known: a+b=8, ab= 16+c2. Find the value of (a-b+c)2002.
Analysis: The value of (a-b+c)2002 cannot be directly derived from the known conditions. We can use (a-b)2=(a+b)2-4ab to determine the relationship between A-b and C, and then calculate the value of (a-b+c) 2002.
Solution: (a-b) 2 = (a+b) 2-4ab = 82-4 (16+C2) =-4c2.
Namely: (a-b)2+4c2=0.
∴a-b=0,c=0。
∴(a-b+c)2002=0。
Example 4 shows that A, B, C and D are positive rational numbers and satisfy a4+b4+C4+D4=4abcd.
Proof: a = b = c = d.
Analysis: From the characteristics of a4+b4+C4+D4=4abcd, we can see that it can be transformed into a complete square, and then find a proof idea.
Proof: ∫a4+B4+C4+D4 = 4 ABCD
∴a4-2a2b2+b4+c4-2c2d2+d4+2a2b2-4abcd+2c2d2=0,
(a2-b2)2+(c2-d2)2+2(ab-cd)2=0 .
a2-b2=0,c2-d2=0,ab-cd=0
And ∵a, b, c and d are positive rational numbers,
∴a=b,c=d。 Substitute ab-cd=0,
A2=c2, that is, A = C. ..
So there is a = b = c = d.
Recommended by 20 19-09-27.
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