Dạng 2: Bài luyện tập 2 có đáp án

$\frac{8}{x^2+4x}+\frac{5}{x+4}-\frac{2}{x}\left(x\ne 0;x\ne -4\right)$

$\begin{array}{l}=\frac{8}{x\left(x+4\right)}+\frac{5x}{x\left(x+4\right)}-\frac{2\left(x+4\right)}{x\left(x+4\right)}\\ \text{=}\frac{-7}{x+4}\end{array}$

$\begin{array}{l}\frac{2}{x-3}-\frac{6}{x^2-9}+\frac{1}{x-3}\left(x\ne \pm 3\right)\\ =\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{6}{\left(x-3\right)\left(x+3\right)}+\frac{1\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\frac{3x+3}{x^2-9}\end{array}$

$\frac{x^2-7x}{x^2-9}.\frac{x^2+6x+9}{x^2-49}$

$\begin{array}{l}=\frac{8}{x\left(x+4\right)}+\frac{5x}{x\left(x+4\right)}-\frac{2\left(x+4\right)}{x\left(x+4\right)}\\ \text{=}\frac{-7}{x+4}\end{array}$

$\begin{array}{l}\frac{2x-1}{3x^2+x}:\frac{8x^3-1}{9x^2+6x+1}\left(x\ne \frac{1}{2};x\ne 0;x\ne \pm \frac{1}{3}\right)\\ =\frac{2x-1}{x\left(3x+1\right)}.\frac{{\left(3x+1\right)}^2}{\left(2x-1\right)\left(4x^2+2x+1\right)}\\ =\frac{3x+1}{x.\left(4x^2+2x+1\right)}\end{array}$

$\begin{array}{l}M-\frac{x}{x+5}=\frac{x+1}{x+5}\\ M=\frac{x+1}{x+5}+\frac{x}{x+5}\\ M=\frac{2x+1}{x+5}\end{array}$

$\begin{array}{l}\frac{x+y}{x-y}.M=\frac{x^2+2xy+y^2}{x^2+xy+y^2}\\ M=\frac{x^2+2xy+y^2}{x^2+xy+y^2}:\frac{x+y}{x-y}\\ M=\frac{x^2-y^2}{x^2+xy+y^2}\end{array}$

$\begin{array}{l}\frac{\frac{m}{m-2}-\frac{m+2}{m}}{\frac{m}{m+2}-\frac{m-2}{m}}\left(x\ne 0;x\ne \pm 2\right)\\ =\frac{\frac{m^2}{\left(m-2\right)m}-\frac{\left(m+2\right)\left(m-2\right)}{m\left(m-2\right)}}{\frac{m^2}{\left(m+2\right)m}-\frac{\left(m+2\right)\left(m-2\right)}{m\left(m+2\right)}}\\ =\frac{m+2}{m-2}\end{array}$

$\begin{array}{l}\frac{\frac{3}{5}-\frac{3}{m+1}}{\frac{16-m^2}{m^2+2m+1}}\left(m\ne 1;m\ne \pm 4\right)\\ =\frac{\frac{3\left(m+1\right)}{5\left(m+1\right)}-\frac{3.5}{5\left(m+1\right)}}{\frac{\left(4-m\right)\left(4+m\right)}{{\left(m+1\right)}^2}}\\ =\frac{-3\left(m+1\right)}{5\left(m+4\right)}\end{array}$

$\begin{array}{l}A=\frac{2}{xy}:{\left(\frac{1}{x}-\frac{1}{y}\right)}^2-\frac{x^2+y^2}{x^2-2xy+y^2}\\ A=\frac{2}{xy}.\frac{x^2.y^2}{{\left(x-y\right)}^2}-\frac{x^2+y^2}{{\left(x-y\right)}^2}\\ A=\frac{{\left(x-y\right)}^2}{{\left(x-y\right)}^2}\\ A=1\end{array}$

$\frac{x^2+8x+16}{x^2-16}=\frac{{\left(x+4\right)}^2}{\left(x+4\right)\left(x-4\right)}=\frac{x+4}{x-4}$

$\begin{array}{l}A=\frac{a^2+2a}{2a+10}+\frac{a-5}{a}+\frac{50-5a}{2a\left(a+5\right)}\\ A=\frac{a\left(a^2+2a\right)}{2a\left(a+5\right)}+\frac{2\left(a-5\right)\left(a+5\right)}{2a\left(a+5\right)}+\frac{50-5a}{2a\left(a+5\right)}\\ A=\frac{a^3+4a^2-5a}{2a\left(a+5\right)}\\ A=\frac{a-1}{2}\end{array}$

$\begin{array}{l}A=\left(1-\frac{4}{x+1}\right):\frac{9-x^2}{x^2+2x+1}\\ A=\left(\frac{x+1}{x+1}-\frac{4}{x+1}\right).\frac{{\left(x+1\right)}^2}{\left(x+3\right)\left(3-x\right)}\\ A=-\frac{x+1}{x+3}\end{array}$

$\left\{\begin{array}{l}x\ne 0\\ x-2\ne 0\end{array}\right.\Rightarrow \left\{\begin{array}{l}x\ne 0\\ x\ne 2\end{array}\right.$

$\begin{array}{l}C=\frac{x^2}{x-2}.\left(\frac{x^2+4}{x}-4\right)+3\\ C=\frac{x^2}{x-2}.\left(\frac{x^2-4x+4}{x}\right)+3\\ C=\frac{x^2}{x-2}.\frac{{\left(x-2\right)}^2}{x}+3\\ C=x.\left(x-2\right)+3\\ C=x^2-2x+3\end{array}$

$\left\{\begin{array}{l}x\ne 0\\ x+2\ne 0\end{array}\right.\Rightarrow \left\{\begin{array}{l}x\ne 0\\ x\ne -2\end{array}\right.$

$\begin{array}{l}D=\frac{{\left(x+2\right)}^2}{x}.\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\\ D=\frac{{\left(x+2\right)}^2}{x}.\frac{x+2-x^2}{x+2}-\frac{x^2+6x+4}{x}\\ D=\frac{\left(x+2\right).\left(x+2-x^2\right)}{x}-\frac{x^2+6x+4}{x}\\ D=\frac{x^3-x^2+4x+4}{x}-\frac{x^2+6x+4}{x}\\ D=\frac{x^3-2x^2-2x}{x}\\ D=x^2-2x-2\end{array}$