Solution: For convenience, we assume that this positive integer is X, and the two complete squares are M 2 and N 2 in turn, so there is
X+ 100 = m 2，X+ 168 = n 2。 Subtract these two expressions to get the final result.
N 2-m 2 = 68, that is, (n+m)(n-m)=68.
We know that (n+m) and (n-m) have the same parity, and 68 is an even number, so (n+m) and (n-m) are even numbers, and 68=2X2X 17, so the only decomposition can only be 34,2. It is easy to get m= 16 and n= 18.
So this positive integer x =162-100 = 256-100 =156.
2. Given (x 2) * (y 2)+x 2+y 2 =10xy-16, find x, y.
Solution: organize the equation into
(xy)^2-8xy+ 16+x^2+y^2-2xy=0
So you got it.
(xy-4)^2+(x-y)^2=0，
Since (xy-4) 2 ≥ 0 and (x-y) 2 ≥ 0, it is obtained from the above equation.
(xy-4) 2 = 0，(x-y) 2 = 0。 Solve it and get it.
Xy=4, X = Y. That is, X 2 = 4, so x=2 or -2. There are also y=2, or -2.
3. Given (a+b) 2-(a-b) 2 = 12,1/3 (a-3)-1/ 3 (2+b) = 5/6, find-1.
Solution: expand (a+b) 2-(a-b) 2 = 12 into a 2+b 2+2ab-(a 2+b 2-2ab) =12.
arrange
4ab= 12, that is, ab=3.
Then expand1/3 (a-3)-1/3 (2+b) = 5/6, as shown below.
2(a-3)-2(2+b)=5。
2a-2b= 15, because a is not equal to 0, so both sides are multiplied by a.
2A 2-2AB = 15A, and substitute ab=3 to get
2a^2- 15a-6=0，
Get a=( 15+ root number 273)/4 or (15- root number 273)/4 by solving.
Algebraic expression = (ab) 2x (1/2a-1/3b)
=3^2X 1/6X(3a-2b)
=3/2X(a+2a-2b)
=3/2X(a+ 15)
Substituting the value of a gives two results: (225+3 root number 73)/8; (225-3 root number 73)/8.
Note: I wish the landlord a happy new year first! Get what you want Your third question may have incorrect data, because I follow my above algorithm and the result is complicated. Please check it carefully.