Đ/A đây:

=\(\frac{2^{15}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}\)

=\(\frac{2^{12}.3^5.\left(2^3-3\right)}{2^{12}.3^5.\left(3+1\right)}\)

                                               cố lên

=\(\frac{5}{4}\)

\(\left|x-\frac{5}{2}\right|+\frac{1}{6}=\frac{2}{3}\)

\(\left|x-\frac{5}{2}\right|=\frac{2}{3}-\frac{1}{6}\)

\(\left|x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-\frac{5}{2}=\frac{1}{2}\\x-\frac{5}{2}=-\frac{1}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=2\end{array}\right.\)

<=>|x-5/2|=2/3-1/6=1/2

=> x-5/2 =1/2   hoặc x-5/2=-1/2

* x-5/2=1/2<=>x=1/2+5/2=3

*x-5/2=-1/2<=>x=-1/2+5/2=2

Lời giải:

$2x+xy-2y=7$

$x(2+y)-2y=7$

$x(2+y)-2(y+2)=3$

$(x-2)(y+2)=3$

Do $x,y$ là số nguyên nên $x-2, y+2$ cũng là số nguyên. Do đó ta có bảng sau:

\(2x+xy-2y=7\)

\(\Rightarrow x\left(2+y\right)-2y-4+4=7\)

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

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

\(\Rightarrow\left(x-2\right);\left(y+2\right)\in\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;-5\right);\left(3;1\right);\left(-1;-3\right);\left(5;-1\right)\right\}\left(x;y\inℤ\right)\)

\(\left|5x-4\right|=\left|x-2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-4=2-x\\5x-4=x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x=6\\4x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

Vậy  \(x\in\left\{1;\frac{1}{2}\right\}\)

\(\left|5x-4\right|=\left|x-2\right|\)

\(\Rightarrow\orbr{\begin{cases}5x-4=x-2\\5x-4=-\left(x-2\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x-x=-2+4\\5x+x=2+4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=2\\6x=6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

Vậy,...........

-x-2/3=-6/7

-x      =-6/7+2/3

-x      =-18/21+14/21

-x      =-4/21

=> x=4/21

=> -x= -(2/3 + 6/7)
=> -x= -32/21

=> x=32/21

mk chưa học đến thông cảm

\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\Leftrightarrow\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)\Leftrightarrow\frac{x^2}{4}=\frac{y^2}{9}\)

Theo t/c dãy tỉ số bằng nhau ;

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{100}{13}\)

\(\Rightarrow x^2=\frac{100}{13}.4\Rightarrow x^2=\frac{400}{13}\Rightarrow x=\hept{\begin{cases}\sqrt{\frac{400}{13}}\\-\sqrt{\frac{400}{13}}\end{cases}}\)

\(\Rightarrow y^2=\frac{100}{13}.9=\frac{900}{13}\Leftrightarrow y=\hept{\begin{cases}\sqrt{\frac{900}{13}}\\-\sqrt{\frac{900}{13}}\end{cases}}\)