1/

\(\frac{x-1}{13}-\frac{2x-13}{15}=\frac{3x-15}{27}-\frac{4x-27}{29}\)

\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)-\left(\frac{2x-13}{15}-1\right)=\left(\frac{3x-15}{27}-1\right)-\left(\frac{4x-27}{29}-1\right)\)

\(\Leftrightarrow\frac{x-14}{13}-\frac{2\left(x-14\right)}{15}=\frac{3\left(x-14\right)}{27}-\frac{4\left(x-14\right)}{29}\)

\(\Leftrightarrow\frac{x-14}{13}-\frac{2\left(x-14\right)}{15}-\frac{3\left(x-14\right)}{27}+\frac{4\left(x-14\right)}{29}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}-\frac{2}{15}-\frac{3}{27}+\frac{4}{29}\right)=0\)

\(\Leftrightarrow x-14=0\)(vì 1/13 -2/15 -3/27 +4/29 khác 0)

\(\Leftrightarrow x=14\)

vậy...................

2/ 

\(a,ĐKXĐ:x\ne\pm2\)

\(b,A=\frac{4}{3x-6}-\frac{x}{x^2-4}\)

          \(=\frac{4}{3\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}\)

           \(=\frac{4\left(x+2\right)-3x}{3\left(x-2\right)\left(x+2\right)}\)

            \(=\frac{x+8}{3\left(x-2\right)\left(x+2\right)}\)

c,với \(x\ne\pm2\)ta có \(A=\frac{x+8}{3\left(x-2\right)\left(x+2\right)}\)

với x=1 thay vào A ta có \(A=\frac{1+8}{3\left(1-2\right)\left(1+2\right)}=\frac{9}{-9}=-1\)

Ta có:

\(Q=\frac{27-2x}{12-x}\)

\(Q=\frac{24-2x+3}{12-x}\)

\(Q=\frac{24-2x}{12-x}+\frac{3}{12-x}\)

\(Q=\frac{2}{1}+\frac{3}{12-x}\)

\(Q=2+\frac{3}{12-x}.\)

a) Để biểu thức Q có giá trị nguyên.

\(\Rightarrow2+\frac{3}{12-x}\) có giá trị nguyên.

\(\Rightarrow\frac{3}{12-x}\) có giá trị nguyên.

\(\Rightarrow3⋮12-x\)

\(\Rightarrow12-x\inƯC\left(3\right)\)

\(\Rightarrow12-x\in\left\{1;-1;3;-3\right\}.\)

\(\Rightarrow\left[{}\begin{matrix}12-x=1\\12-x=-1\\12-x=3\\12-x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12-1\\x=12+1\\x=12-3\\x=12+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=11\left(TM\right)\\x=13\left(TM\right)\\x=9\left(TM\right)\\x=15\left(TM\right)\end{matrix}\right.\)

Vậy \(x\in\left\{11;13;9;15\right\}\) thì Q có giá trị nguyên.

Chúc bạn học tốt!

Đề sai ạ ! Sửa lại nhé : 

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)

\(\Leftrightarrow A=\frac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\frac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right)\)

\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{-x^2+3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}.\frac{3\left(x-3\right)\left(x+3\right)}{-x^2+3x-9}\)

\(\Leftrightarrow A=\frac{-\left(x+3\right)}{x}\)

b) Để \(A\inℤ\)

\(\Leftrightarrow-\left(x+3\right)⋮x\)

\(\Leftrightarrow-x-3⋮x\)

\(\Leftrightarrow3⋮x\)

\(\Leftrightarrow x\inƯ\left(3\right)\)

Vậy để \(A\inℤ\Leftrightarrow x\inƯ\left(3\right)\)(\(x\neℤ\))

Bạn sửa cho mik dòng cuối :

\(x\ne Z\)thành \(x\notin Z\)nhé !

Vâng!

\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)

        \(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)

         \(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)

        \(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)

       \(=-3\)

\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)

     \(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

    \(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)

    \(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

b, Ta có \(B< A\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)

\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)

\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)

\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)

Vậy ...