Giải bài 1 trang 63 SGK Toán 7 tập 2 - Cánh diều

Đề bài

Tính:

a) \(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3}\);                                  b) \({y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25)\);

c) \((2{x^2} + x + 4)({x^2} - x - 1)\);                                            d) \((3x - 4)(2x + 1) - (x - 2)(6x + 3)\).

Lời giải chi tiết

a) \(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);                                                   

b) \(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);

c) \(\begin{array}{l}(2{x^2} + x + 4)({x^2} - x - 1) = 2{x^2}({x^2} - x - 1) + x({x^2} - x - 1) + 4({x^2} - x - 1)\\ = 2{x^4} - 2{x^3} - 2{x^2} + {x^3} - {x^2} - x + 4{x^2} - 4x - 4 = 2{x^4} - {x^3} + {x^2} - 5x - 4\end{array}\);                                                               

d) \(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) = 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).