a) \(\frac{1-x}{x+4}=\frac{5-4-x}{x+4}=\frac{5}{x+4}-1\inℤ\Leftrightarrow\frac{5}{x+4}\inℤ\)

mà \(x\inℤ\Rightarrow x+4\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\)

\(\Leftrightarrow x\in\left\{-9,-5,-3,1\right\}\)

b) \(\frac{11-2x}{x-5}=\frac{1+10-2x}{x-5}=\frac{1}{x-5}-2\inℤ\Leftrightarrow\frac{1}{x-5}\inℤ\)

mà \(x\inℤ\Rightarrow x-5\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{4,6\right\}\)

c) \(\frac{x+1}{2x+1}\inℤ\Rightarrow\frac{2\left(x+1\right)}{2x+1}=\frac{2x+1+1}{2x+1}=1+\frac{1}{2x+1}\inℤ\Leftrightarrow\frac{1}{2x+1}\inℤ\)

mà \(x\inℤ\Rightarrow2x+1\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{-1,0\right\}\).

Thử lại đều thỏa mãn. 

4(x - 3) + 12 = 6

4( x - 3)        = 6 - 12

4(x - 3)         = -6

   x - 3        = \(\frac{-3}{2}\)

    \(x=\frac{-3}{2}+3=\frac{3}{2}\)

\(4\left(x-3\right)+12=6\)

\(\Leftrightarrow4\left(x-3\right)=-6\)

\(\Leftrightarrow4\left(x-3\right)=-6\)

\(\Leftrightarrow x-3=-\frac{3}{2}\)

\(\Leftrightarrow x=\frac{3}{2}\)

#H

a) \(A=\frac{3x+2}{x-3}\)

Bạn đọc tự làm. 

b) \(A=\frac{3x+2}{x-3}=\frac{3x-9+11}{x-3}=3+\frac{11}{x-3}\inℤ\Leftrightarrow\frac{11}{x-3}\inℤ\)

mà \(x\inℤ\Rightarrow x-3\inƯ\left(11\right)=\left\{-11,-1,1,11\right\}\)

\(\Leftrightarrow x\in\left\{-8,1,4,14\right\}\)

c)  \(A=\frac{x^2+3x-7}{x+3}=\frac{x\left(x+3\right)-7}{x+3}=x-\frac{7}{x+3}\inℤ\Leftrightarrow\frac{7}{x+3}\inℤ\)

mà \(x\inℤ\Rightarrow x+3\inƯ\left(7\right)=\left\{-7,-1,1,7\right\}\)

\(\Leftrightarrow x\in\left\{-10,-4,-2,4\right\}\)

d) \(x\in\left\{4\right\}\).

5\(x\) - 16 = 40 + \(x\)

5\(x\) - \(x\) = 40 + 16

 4\(x\)     = 56

    \(x=56:4\)

    \(x=14\)

Vậy \(x=14\)

b; 4\(x\) - 10 = 15 - \(x\)

   4\(x\) + \(x\) = 15 +10

      5\(x=25\)

       \(x=25:5\)

        \(x=5\)

Vậy \(x=5\)

c; -12 + \(x\) = 5\(x-20\)

     5\(x\) = - 12 + \(x\) + 20

     5\(x\) - \(x\) = - 12+ 20

       4\(x\) = 8

      \(x=\dfrac{8}{4}\)

     \(x=2\)

Vậy \(x=2\)

d; \(x+15\) = 7 - 6\(x\)

   6\(x\) = 7 - \(x\) - 15

   6\(x\) + \(x\) = 7 - 15

   7\(x\) =  - 8

     \(x=-\dfrac{8}{7}\)

Vậy \(x=-\dfrac{8}{7}\) 

\(1,a^2-2a+1-b^2\)

\(=\left(a^2-2a+1\right)-b^2\)

\(=\left(a-1\right)^2-b^2\)

\(=\left(a-1-b\right)\left(a-1+b\right)\)       Khai triển thành hằng đẳng thức số 3 e  nhé.

\(2,x^2+2xy+y^2-81\)

\(=\left(x^2+2xy+y^2\right)-81\)

\(=\left(x+y\right)^2-9^2\)

\(=\left(x+y-9\right)\left(x+y+9\right)\)Cái này cũng HĐT số 3 nè

\(3,x^2+6y-9-y^2\)

\(=-\left(y^2-6y+9\right)+x^2\)

\(=-\left(y-3\right)^2+x^2\)

\(=x^2-\left(y-3\right)^2\)

\(=\left(x-y-3\right)\left(x-y+3\right)\)

\(5,4x^2+y^2-9-4xy\)

\(=\left(4x^2-4xy+y^2\right)-9\)

\(=\left(2x-y\right)^2-3^2\)

\(=\left(2x-y-3\right)\left(2x-y+3\right)\)

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