Đề kiểm tra 15 phút - Đề số 5 - Bài 6 - Chương 1 - Đại số 8

Bài 1. Phân tích thành nhân tử:

a) \(a\left( {b - 3} \right) + \left( {3 - b} \right) - b\left( {3 - b} \right)\)

b) \(15{a^2}b\left( {{x^2} - y} \right)20a{b^3}\left( {{x^2} - y} \right) + 25ab\left( {y - {x^2}} \right)\)

c) \(5{\left( {a - b} \right)^2} - \left( {a + b} \right)\left( {b - a} \right).\)

Bài 2. Tìm x, biết:

a) \(x\left( {x - 4} \right) = 2x - 8\)

b) \(\left( {2x + 3} \right)\left( {x - 1} \right) + \left( {2x - 3} \right)\left( {1 - x} \right) = 0.\)

Bài 1.

a) \(a\left( {b - 3} \right) + \left( {3 - b} \right) - b\left( {3 - b} \right) \)

\(=  - a\left( {3 - b} \right) + \left( {3 - b} \right) - 3\left( {3 - b} \right)\)

\( = \left( {3 - b} \right)\left( { - a + 1 - b} \right).\)

b) \(15{a^2}b\left( {{x^2} - y} \right) - 20a{b^3}\left( {{x^2} - y} \right) + 25ab\left( {y - {x^2}} \right)\)

\( = 15{a^2}b\left( {{x^2} - y} \right) - 20a{b^2}\left( {{x^2} - y} \right) - 25ab\left( {{x^2} - y} \right)\)

\( = \left( {{x^2} - y} \right)\left( {15{a^2}b - 20a{b^2} - 25ab} \right)\)

\(= \left( {{x^2} - y} \right).5ab\left( {3a - 4b - 5} \right).\)

c) \(5{\left( {a - b} \right)^2} - \left( {a + b} \right)\left( {b - a} \right) \)

\(= 5{\left( {a - b} \right)^2} + \left( {a + b} \right)\left( {a - b} \right)\)

\( = \left( {a - b} \right)\left[ {5\left( {a - b} \right) + \left( {a + b} \right)} \right]\)

\(= 2\left( {a - b} \right)\left( {3a - 2b} \right).\)

Bài 2.

a) \(x\left( {x - 4} \right) = 2x - 8\)

\(\Rightarrow x\left( {x - 4} \right) = 2\left( {x - 4} \right)\)

\( \Rightarrow x\left( {x - 4} \right) - 2\left( {x - 4} \right) = 0 \)

\(\Rightarrow \left( {x - 4} \right)\left( {x - 2} \right) = 0\)

\( \Rightarrow x - 4 = 0\)  hoặc \(x - 2 = 0\)

\(\Rightarrow x = 4\) hoặc \(x = 2.\)

b) \(\left( {2x + 3} \right)\left( {x - 1} \right) + \left( {2x - 3} \right)\left( {1 - x} \right) = 0\)

\( \Rightarrow \left( {2x + 3} \right)\left( {x - 1} \right) - \left( {2x - 3} \right)\left( {x - 1} \right) = 0\)

\( \Rightarrow \left( {x - 1} \right)\left[ {\left( {2x + 3} \right) - \left( {2x - 3} \right)} \right] = 0\)

\( \Rightarrow 6\left( {x - 1} \right) = 0\)

\(\Rightarrow x - 1 = 0 \Rightarrow x = 1\)