Tính: (3/4x^3):(−1/2x^2)

Tính:

Giải sách bài tập Toán lớp 7 Bài 5: Phép chia đa thức một biến

Bài 41 trang 53 sách bài tập Toán lớp 7 Tập 2: Tính:

a) $\left(\frac{3}{4}{x}^3\right):\left(-\frac{1}{2}{x}^2\right)$;

b) (5xⁿ) : (4x²) (n ∈ ℕ, n ≥ 2);

c) $\left(x^3-3x^2+6x\right):\left(-\frac{1}{3}x\right);$

d) $\left(x+\frac{1}{3}{x}^2+\frac{7}{2}{x}^3\right):\left(5x\right).$

Lời giải:

a) $\left(\frac{3}{4}{x}^3\right):\left(-\frac{1}{2}{x}^2\right)$$=\left(\frac{3}{4}:\left(-\frac{1}{2}\right)\right).\left(x^3:x^2\right)$

$=\frac{3}{4}.\frac{-2}{1}.x^{3-2}=-\frac{3}{2}x$;

b) (5xⁿ) : (4x²) = (5 : 4) . (xⁿ – x²) = $\frac{5}{4}$x^{n – 2} (n ∈ℕ, n ≥ 2);

c) $\left(x^3-3x^2+6x\right):\left(-\frac{1}{3}x\right)$

$=\left(x^3:\left(-\frac{1}{3}x\right)\right)-\left(3x^2:\left(-\frac{1}{3}x\right)\right)+\left(6x:\left(-\frac{1}{3}x\right)\right)$

$=\left(1:\left(-\frac{1}{3}\right)\right).\left(x^3:x\right)-\left(3:\left(-\frac{1}{3}\right)\right).\left(x^2:x\right)+\left(6:\left(-\frac{1}{3}\right)\right).\left(x:x\right)$

= 1 . (–3) . x^{3 – 1} – 3 . (–3) . x^{2 – 1} + 6 . (–3) . x^{1 – 1}

= –3x² + 9x – 18;

d) $\left(x+\frac{1}{3}{x}^2+\frac{7}{2}{x}^3\right):\left(5x\right)$

$=\left(x:5x\right)+\left(\frac{1}{3}{x}^2:5x\right)+\left(\frac{7}{2}{x}^3:5x\right)$

$=\left(1:5\right).\left(x:x\right)+\left(\frac{1}{3}:5\right).\left(x^2:x\right)+\left(\frac{7}{2}:5\right).\left(x^3:x\right)$

$=\frac{1}{5}.x^{1-1}+\frac{1}{3}.\frac{1}{5}.x^{2-1}+\frac{7}{2}.\frac{1}{5}.x^{3-1}$

$=\frac{1}{5}{x}^0+\frac{1}{15}{x}^1+\frac{7}{10}{x}^2$

$=\frac{1}{5}+\frac{1}{15}x+\frac{7}{10}{x}^2.$