Câu 1:
a: \(x^2-xy-y+4=0\)
=>\(x^2-1-xy-y+5=0\)
=>\(\left(x-1\right)\left(x+1\right)-y\left(x+1\right)=-5\)
=>\(\left(x+1\right)\left(x-1-y\right)=-5\)
=>\(\left(x+1\right)\left(x-1-y\right)=1\cdot\left(-5\right)=\left(-5\right)\cdot1=\left(-1\right)\cdot5=5\cdot\left(-1\right)\)
=>\(\left(x+1;x-1-y\right)\in\left\{\left(1;-5\right);\left(-5;1\right);\left(-1;5\right);\left(5;-1\right)\right\}\)
=>\(\left(x;x-1-y\right)\in\left\{\left(0;-5\right);\left(-6;1\right);\left(-2;5\right);\left(4;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;4\right);\left(-6;-8\right);\left(-2;-8\right);\left(4;4\right)\right\}\)
b: \(A=n^3-n^2-n-2\)
\(=n^3-2n^2+n^2-2n+n-2\)
\(=\left(n-2\right)\left(n^2+n+1\right)\)
TH1: n=3
=>A=(3-2)(3^2+3+1)=9+3+1=13(nhận)
TH2: n=3k+1
\(A=\left(3k+1-2\right)\left[\left(3k+1\right)^2+3k+1+1\right]\)
\(=\left(3k-1\right)\left(9k^2+6k+1+3k+2\right)\)
\(=\left(3k-1\right)\left(9k^2+9k+3\right)\)
\(=3\cdot\left(3k-1\right)\left(3k^2+3k+1\right)⋮3\)
=>Loại
TH2: n=3k+2
\(A=\left(3k+2-2\right)\left[\left(3k+2\right)^2+3k+2+1\right]\)
\(=3k\cdot\left(9k^2+12k+4+3k+3\right)\)
\(=3k\left(9k^2+15k+7\right)⋮3\)
=>Loại
Câu 2:
a: \(A=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{x}{x^2-2x+1}-\dfrac{1}{x^2-1}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x^2-1\right)}{x^2+1}\cdot\left(\dfrac{x}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x\left(x+1\right)-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{x^2+x-x+1}{x-1}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x-1}=-1\)