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

A \(x \ne 1\)

B \(x \ne  - 1\)      

C \(x \ne 0\)         

D \(\forall x \in R\)

A \(\frac{{4x - 8}}{x}\)

B \(4x - 8\)          

C \(2x - 4\)          

D \(2x - 2\)

A \(\frac{3}{2}\,dm\)

B \(1\,dm\)

C \(\sqrt 2 \,dm\)

D \(2\,dm\)

A \(5\,cm\)           

B \(1\,dm\)

C \(\sqrt 2 \,dm\)

D \(2\,dm\)

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

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

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

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

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

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

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

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

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

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

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

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