\(\dfrac{x-5}{6}=\dfrac{x+1}{3}\\ \Rightarrow\dfrac{x-5}{6}=\dfrac{2\left(x+1\right)}{6}\\ \Rightarrow x-5=2\left(x+1\right)\\ \Rightarrow x-5=2x+2\\ \Rightarrow2x+2-x+5=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\)

\(\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}+\dfrac{2}{13}}{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{9}-\dfrac{3}{11}+\dfrac{3}{13}}+\dfrac{15151515}{45454545}\\ =\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{1}{3}\\ =\dfrac{2}{3}+\dfrac{1}{3}\\ =\dfrac{3}{3}\\ =1\)

Chọn A

\(A+B=5xyz-6^2+8xy+3x^2+2xyz-8xy+y^2\\ =7xyz+3x^2+y^2-36\)

a, ĐKXĐ:\(\left\{{}\begin{matrix}x^2-1\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm1\\x\ne-1\\x\ne1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)

b, \(P=\dfrac{2x^2}{x^2-1}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\)

\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2-x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x^2+x}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow P=\dfrac{2x^2+x^2-x-x^2-x}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow P=\dfrac{2x^2-2x}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow P=\dfrac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow P=\dfrac{2x}{x+1}\)

c, Thay x=2 vào P ta có:

\(P=\dfrac{2x}{x+1}=\dfrac{2.2}{2+1}=\dfrac{4}{3}\)

C

Vì x,y nguyên nên \(\left\{{}\begin{matrix}2x+1,y-2\in Z\\2x+1,y-2\inƯ\left(-5\right)\end{matrix}\right.\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(-3;3\right);\left(-1;7\right);\left(0;-3\right);\left(2;1\right)\right\}\)

`6xxy=5/8:3/4`

`6xxy=5/8xx4/3`

`6xxy=5/6`

`y=5/6:6`

`y=5/6xx1/6`

`y=5/36` 

đề phải là x,y nguyên nếu ko có thì đề sai

`3xx4/15-2xx4/15-5xx4/15`

`=4/15xx(3-2-5)`

`=4/15xx(-4)`

`=-16/15`