Processing math: 100%

Đề thi HK1 Toán 8 - Sở GD&ĐT Yên Khánh - Hà Nội -...

$x \ne 1$

$x \ne - 1$

$x \ne 0$

$\forall x \in R$

$\frac{{4x - 8}}{x}$

$4x - 8$

$2x - 4$

$2x - 2$

$\frac{3}{2}\,dm$

$1\,dm$

$\sqrt 2 \,dm$

$2\,dm$

$5\,cm$

$1\,dm$

$\sqrt 2 \,dm$

$2\,dm$

$\begin{array}{l}a)\,\,x\left( {x + 2} \right)\\b)\,\,\left( {x - 1} \right)\left( {{x^2} - 4x + 1} \right)\end{array}$

$\begin{array}{l}a)\,\,x\left( {x + 2} \right)\\b)\,\,\left( {x - 1} \right){\left( {x - 2} \right)^2}\end{array}$

$\begin{array}{l}a)\,\,x\left( {x - 2} \right)\\b)\,\,\left( {x + 1} \right){\left( {x - 2} \right)^2}\end{array}$

$\begin{array}{l}a)\,\,x\left( {x - 2} \right)\\b)\,\,\left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\end{array}$

$\begin{array}{l}a)\,\,{x^2} - 2x + 3\\b)\,\,\frac{{4x{{\left( {x - 2} \right)}^2}}}{{{{\left( {x + 5} \right)}^2}\left( {x - 5} \right)}}\end{array}$

$\begin{array}{l}a)\,\,{x^2} - 2x - 3\\b)\,\,\frac{{4x{{\left( {x - 2} \right)}^2}}}{{{{\left( {x - 5} \right)}^2}\left( {x + 5} \right)}}\end{array}$

$\begin{array}{l}a)\,\,{x^2} - 2x + 3\\b)\,\,\frac{{4x{{\left( {x - 2} \right)}^2}}}{{{{\left( {x + 5} \right)}^2}\left( {5 - x} \right)}}\end{array}$

$\begin{array}{l}a)\,\,{x^2} - 2x - 3\\b)\,\,\frac{{4x{{\left( {x - 2} \right)}^2}}}{{{{\left( {x - 5} \right)}^2}\left( {x + 5} \right)}}\end{array}$

$\begin{array}{l}a)\,\,\left\{ \begin{array}{l}x \ne 0\\x \ne 1\end{array} \right.\\b)\,\,A = \frac{{x - 1}}{x}\end{array}$

$\begin{array}{l}a)\,\,x \ne 0\\b)\,\,A = \frac{{x - 1}}{{x + 1}}\end{array}$

$\begin{array}{l}a)\,\,x \ne 1\\b)\,\,A = \frac{{x - 1}}{{x\left( {x + 1} \right)}}\end{array}$

$\begin{array}{l}a)\,\,\left\{ \begin{array}{l}x \ne 0\\x \ne - 1\end{array} \right.\\b)\,\,A = \frac{{x + 1}}{x}\end{array}$