Dạng 2. Chứng minh một phân thức luôn có nghĩa có đáp án

$\frac{7}{x^2+5}$ $x^2\ge 0,\forall x$$\Rightarrow x^2+5\ge 5,\forall x$

$\frac{6-x}{{\left(x+1\right)}^2+4}$ ${\left(x+1\right)}^2\ge 0,\forall x$$\Rightarrow {\left(x+1\right)}^2+4\ge 4,\forall x$

$\frac{8-x^2}{x^2+2x+9}$ $x^2+2x+9={\left(x+1\right)}^2+8\ge 8,\forall x$

$\frac{-2x-11}{-x^2+4x-5}$ $-x^2+4x-5=-{\left(x-2\right)}^2-1\le -1;\forall x$

$\frac{3y}{4}=\frac{6xy}{8x}$ $3y.8x=24xy$ $4.6xy=24xy$

$\frac{x+y}{3x}=\frac{3x{\left(x+y\right)}^2}{9x^2\left(x+y\right)}$

$\left(x+y\right).9x^2\left(x+y\right)=9x^2{\left(x+y\right)}^2$

$3x.3x{\left(x+y\right)}^2=9x^2{\left(x+y\right)}^2$

$\frac{2x^2+4x}{{\left(x+2\right)}^2}=\frac{2x}{x+2}$

$\left(2x^2+4x\right)\left(x+2\right)=2x^3+8x^2+8x$

${\left(x+2\right)}^2.2x=2x^4+8x^3+8x$

$\frac{x+1}{x+3}=\frac{x^2+4x+3}{x^2+6x+9}$

$\left(x+1\right)\left(x^2+6x+9\right)=\left(x+1\right){\left(x+3\right)}^2$

$\left(x+3\right)\left(x^2+4x+3\right)={\left(x+3\right)}^2\left(x+1\right)$

$\frac{x-2}{-x}=\frac{8-x^3}{x\left(x^2+2x+4\right)}\frac{x-2}{-x}=\frac{8-x^3}{x\left(x^2+2x+4\right)}$

$\left(x-2\right).x.\left(x^2+2x+4\right)=x\left(x-2\right)\left(x^2+2x+4\right)$

$-x.\left(8-x^3\right)=-x\left(2-x\right)\left(x^2+2x+4\right)=x\left(x-2\right)\left(x^2+2x+4\right)$

$\frac{x^2+y^2+2xy-1}{x^2-y^2+2x+1}=\frac{x+y-1}{x-y+1}$

$\left(x^2+y^2+2xy-1\right)\left(x-y+1\right)=\left(x+y-1\right)\left(x+y+1\right)\left(x-y+1\right)$

$\left(x^2-y^2+2x+1\right)\left(x+y-1\right)=\left(x+1-y\right)\left(x+1+y\right)\left(x+y-1\right)$