If the algebraic formula 9x 18x 5 is a perfect square number, find the value of the integer x

9x -18x+5 is the perfect square number, let 9x -18x+5=y 2

then (3x-3) 2-y 2=4

3x-3+y)(3x-3-y)=4

Because 4 can be decomposed into 1x4, 4x1, 2x2, (-1) x(-4), (4) x(-1), (2) x(-2), a total of 6 kinds of equations, so 6 systems of equations can be obtained: the first system of equations: 3x-3+y=1, 3x-3-y=4, the solution is x=11 6, which does not meet the topic;

The second system of equations: 3x-3+y=4, 3x-3-y=1, solved x=11 6, does not meet the meaning of the topic;

The third system of equations: 3x-3+y=2, 3x-3-y=2, solved x=10 6, does not meet the meaning of the problem;

The fourth system of equations: 3x-3+y=-1, 3x-3-y=-4, solved x=1 6, does not meet the meaning of the topic;

The fifth system of equations: 3x-3+y=-4, 3x-3-y=-1, solved x=1 6, does not meet the meaning of the topic;

The 6th system of equations: 3x-3+y=-2, 3x-3-y=-2, the solution is x=1 3, which does not meet the meaning of the topic;

Therefore, the absence of an integer x makes the algebraic formula 9x -18x+5 a perfectly squared number.

4+ 3/(x-1)

4 is an integer, and the absolute value of x-1 is only a divisor of 3, i.e., x-1 = plus or minus 3, or x-1 = plus or minus 1

x = 4 or 2 or 0 or -2

14x+5-21x²=-9

So 21x -14x = 14

Divide by 73x -2x=2 on both sides

Multiply both sides by 26x -4x = 4

So 6x -4x-5 = 4 - 5 = -1

The square of fraction (6x+18) (9-x) is a positive integer, i.e., fraction (6x+18) (9-x) is a non-zero integer, i.e., x≠-3 and x≠9;

Let the non-zero integer k=(6x+18) (9-x) with wild oak.

Therefore, k = (6x-54+72) (9-x)=-6+72 (9-x), that is, 9-x can be;

So x, -3, -9, -15, -27, -63 and x≠-3, so all possible values are:

Pei Pan, -9, -15, -27, -63

3x²-9x+8

3(x²-3x+9/4-9/4)+8

3(x²-3x+9/4)-27/4+8

3(x-3/2)²+5/4≥5/4>0

So the value is always positive.

3x²-9x+8

3(x-3/2)²+5/4

So x=3 2, and the minimum value is 5 4

Solve the equation first. Reduced to 2x squared - x-1=0

Continue simplifying (2x+1)(x-1)=0

then x=-1 2 or x 1

substitution, when x 1 2.

9 1 4 6 ( 1 2) 5 1 4 when x 1 is 9 5 4 0

(6x+18)/(9-x²)

6(x+3)/[(3+x)(3-x)]

6/(3-x)

Since the fraction 6x+18 9-x squared is a positive integer, 6 (3-x)=6

x=2② 6/(3-x)=3

x=1③ 6/(3-x)=2

x=0④ 6(3-x)=1

x=-3

Let y=6x 2+9y+7 be 2x 2+3y+7=8, and get 2x 2+3y=1

Sort out y=6x 2+9y+7=3(2x 2+3y)+8, and substitute 2x 2+3y=1 into 3(2x 2+3y)+8 to get y=11.