Đề bài - luyện tập 1 trang 20 tài liệu dạy – học toán 8 tập 1

\(\eqalign{ & a)\,\,2x\left( {x - 5} \right) + \left( {x - 2} \right)\left( {x + 3} \right) \cr & = 2{x^2} - 10x + {x^2} + 3x - 2x - 6 \cr & = 3{x^2} - 9x - 6 \cr & b)\,\,\left( {2x - 5} \right)\left( {1 - x} \right) - \left( {x - 3} \right)\left( { - 2x} \right) \cr & = \left( {2x - 2{x^2} - 5 + 5x} \right) - \left( { - 2{x^2} + 6x} \right) \cr & = 2x - 2{x^2} - 5 + 5x + 2{x^2} - 6x \cr & = x - 5 \cr & c)\,\,\left( {4x - 3} \right)\left( {4x - 3} \right) - \left( {3x + 2} \right)\left( {3x - 2} \right) \cr & = 16{x^2} - 24x + 9 - \left( {9{x^2} - 4} \right) \cr & = 16{x^2} - 24x + 9 - 9{x^2} + 4 \cr & = 7{x^2} - 24x + 13 \cr & d)\,\,\left( {x - 2} \right)\left( {x - 1} \right)\left( {x + 3} \right) - {x^2}\left( {x - 5} \right) \cr & = \left( {x - 2} \right)\left( {{x^2} + 3x - x - 3} \right) - {x^2}\left( {x - 5} \right) \cr & = \left( {x - 2} \right)\left( {{x^2} + 2x - 3} \right) - {x^2}\left( {x - 5} \right) \cr & = {x^3} + 2{x^2} - 3x - 2{x^2} - 4x + 6 - {x^3} + 5{x^2} \cr & = 5{x^2} - 7x + 6 \cr} \)

Đề bài

Thực hiện các phép tính:

a) \(2x(x - 5) + (x - 2)(x + 3)\) ;

b) \((2x - 5)(1 - x) - (x - 3)( - 2x)\) ;

c) \((4x - 3)(4x - 3) - (3x + 2)(3x - 2)\) ;

d) \((x - 2)(x - 1)(x + 3) - {x^2}(x - 5)\) .

Lời giải chi tiết

\(\eqalign{ & a)\,\,2x\left( {x - 5} \right) + \left( {x - 2} \right)\left( {x + 3} \right) \cr & = 2{x^2} - 10x + {x^2} + 3x - 2x - 6 \cr & = 3{x^2} - 9x - 6 \cr & b)\,\,\left( {2x - 5} \right)\left( {1 - x} \right) - \left( {x - 3} \right)\left( { - 2x} \right) \cr & = \left( {2x - 2{x^2} - 5 + 5x} \right) - \left( { - 2{x^2} + 6x} \right) \cr & = 2x - 2{x^2} - 5 + 5x + 2{x^2} - 6x \cr & = x - 5 \cr & c)\,\,\left( {4x - 3} \right)\left( {4x - 3} \right) - \left( {3x + 2} \right)\left( {3x - 2} \right) \cr & = 16{x^2} - 24x + 9 - \left( {9{x^2} - 4} \right) \cr & = 16{x^2} - 24x + 9 - 9{x^2} + 4 \cr & = 7{x^2} - 24x + 13 \cr & d)\,\,\left( {x - 2} \right)\left( {x - 1} \right)\left( {x + 3} \right) - {x^2}\left( {x - 5} \right) \cr & = \left( {x - 2} \right)\left( {{x^2} + 3x - x - 3} \right) - {x^2}\left( {x - 5} \right) \cr & = \left( {x - 2} \right)\left( {{x^2} + 2x - 3} \right) - {x^2}\left( {x - 5} \right) \cr & = {x^3} + 2{x^2} - 3x - 2{x^2} - 4x + 6 - {x^3} + 5{x^2} \cr & = 5{x^2} - 7x + 6 \cr} \)