Giải bài 7.24 trang 38 SGK Toán 7 tập 2 – Kết nối tri thức

Đề bài

Rút gọn các biểu thức sau:

a) 4x²(5x² + 3) – 6x(3x³ – 2x + 1) – 5x³ (2x – 1)

b) \(\dfrac{3}{2}x\left( {{x^2} – \dfrac{2}{3}x + 2} \right) – \dfrac{5}{3}{x^2}(x + \dfrac{6}{5})\)

Lời giải chi tiết

a) 4x²(5x² + 3) – 6x(3x³ – 2x + 1) – 5x³ (2x – 1)

= 4x² . 5x² + 4x² . 3 – [6x . 3x³ + 6x . (-2x) + 6x . 1] – [5x³ . 2x + 5x³ . (-1)]

= 20x⁴ + 12x² – (18x⁴ – 12x² + 6x) – (10x⁴ – 5x³)

= 20x⁴ + 12x² – 18x⁴ + 12x² – 6x – 10x⁴ + 5x³

= (20x⁴ – 18x⁴ – 10x⁴ ) + 5x³ + (12x² + 12x² ) – 6x

= -8x⁴ + 5x³ + 24x² – 6x

\(\begin{array}{l}b)\dfrac{3}{2}x\left( {{x^2} – \dfrac{2}{3}x + 2} \right) – \dfrac{5}{3}{x^2}(x + \dfrac{6}{5})\\ = \dfrac{3}{2}x.{x^2} + \dfrac{3}{2}x.( – \dfrac{2}{3}x) + \dfrac{3}{2}x.2 – (\dfrac{5}{3}{x^2}.x + \dfrac{5}{3}{x^2}.\dfrac{6}{5})\\ = \dfrac{3}{2}{x^3} – {x^2} + 3x – (\dfrac{5}{3}{x^3} + 2{x^2})\\ = \dfrac{3}{2}{x^3} – {x^2} + 3x – \dfrac{5}{3}{x^3} – 2{x^2}\\ = (\dfrac{3}{2}{x^3} – \dfrac{5}{3}{x^3}) + ( – {x^2} – 2{x^2}) + 3x\\ = \dfrac{{ – 1}}{6}{x^3} – 3{x^2} + 3x\end{array}\)