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Several math problems in senior two (there must be a process)
1, a positive integer, if you add 100 books, a complete square number; If you add 168, it is a complete square number. Find this positive integer.

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.