Question

xsquqre+1÷x_2_3x+3÷x

Answer 1

Given,

$x^{2}+\frac{1}{x^{2}}-2-3x+\frac{3}{x}\\\;\\=x^{2}+\frac{1}{x^{2}}-2x\times\frac{1}{x}-3(x-\frac{1}{x}})\\\;\\=[x^{2}+\frac{1}{x^{2}}-2x\times\frac{1}{x}]-3(x-\frac{1}{x})\\\;\\=(x-\frac{1}{x})^{2}-3(x-\frac{1}{x})\\\;\\=(x-\frac{1}{x})(x-\frac{1}{x}-3)\\\;\\Note:\;a^{2}+b^{2}+2ab=(a+b)^{2}$