Đề kiểm tra 15 phút - Đề số 1 - Bài 7 - Chương 2 - Đại số 8

1. \(A = {{a - b} \over {{a^2} + ab}}.{{{b^2} + ab} \over {{{\left( {a - b} \right)}^2}}}\)

2. \(B = \left( {{{{x^2} + {y^2}} \over {{x^2} - {y^2}}} - 1} \right).{{x - y} \over {2y}}.\)

3. \(C = {{{a^2} + 2ab + {b^2}} \over {{a^2} + ab + {b^2}}}.{{{a^3} - {b^3}} \over {7a + 7b}}\)

4. \(D = \left( {{{a + 1} \over {2a - 2}} - {1 \over {2{a^2} - 2}}} \right).{{2a + 2} \over {a + 2}}.\)

1.

\(A = {{a - b} \over {{a^2} + ab}}.{{{b^2} + ab} \over {{{\left( {a - b} \right)}^2}}} = {{\left( {a - b} \right)b\left( {a + b} \right)} \over {a\left( {a + b} \right){{\left( {a - b} \right)}^2}}} \)\(\;= {b \over {a\left( {a - b} \right)}}.\)

2.

\(B = \left[ {{{{x^2} + {y^2} - \left( {{x^2} - {y^2}} \right)} \over {{x^2} - {y^2}}}} \right].{{x - y} \over {2y}} = {{2{y^2}\left( {x - y} \right)} \over {\left( {{x^2} - {y^2}} \right)2y}}\)\(\; = {y \over {x + y}}.\)

3.

\(C = {{{{\left( {a + b} \right)}^2}\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \over {7\left( {{a^2} + ab + {b^2}} \right)\left( {a + b} \right)}} = {{{a^2} - {b^2}} \over 7}.\)

4.Ta có: \({{a + 1} \over {2a - 2}} - {1 \over {2{a^2} - 2}} = {{{{\left( {a + 1} \right)}^2} - 1} \over {2\left( {{a^2} - 1} \right)}} = {{{a^2} + 2a + 1 - 1} \over {2\left( {{a^2} - 1} \right)}} \)\(\;= {{a\left( {a + 2} \right)} \over {2\left( {{a^2} - 1} \right)}}\)

Vậy \(D = {{a\left( {a + 2} \right)} \over {2\left( {{a^2} - 1} \right)}}.{{2\left( {a + 1} \right)} \over {a + 2}} = {a \over {a - 1}}.\)