Tính: a) ( 6x^2 – 2x + 1) : ( 3x – 1 ) ; b) ( 27x^3 + x^2 – x + 1) : ( –2x + 1) ; c) (8x^3 + 2x^2 + x) : (2x^3 + x + 1)

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

Câu 4 trang 59 vở bài tập Toán lớp 7 Tập 2:

a) ( 6x² – 2x + 1) : ( 3x – 1 )

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b) ( 27x³ + x² – x + 1) : ( –2x + 1)

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c) (8x³ + 2x² + x) : (2x³ + x + 1)

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d) ( 3x⁴ + 8x³ – 2x² + x + 1) : ( 3x + 1) ........................................................................................................................

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Lời giải:

a) ( 6x² – 2x + 1) : ( 3x – 1 )

Vậy (6x² – 2x + 1) : (3x – 1) = 2x ( dư 1).

b) ( 27x³ + x² – x + 1) : (–2x + 1)

Vậy ( 27x³ + x² – x + 1) : ( –2x + 1) = $\frac{-27}{2}{\text{x}}^2-\frac{29}{4}\text{x}-\frac{25}{8}$ ( dư $\frac{33}{8}$).

c) (8x³ + 2x² + x) : (2x³ + x + 1)

Vậy (8x³ + 2x² + x) : (2x³ + x + 1) = 4 (dư 2x² – 3x).

d) ( 3x⁴ + 8x³ – 2x² + x + 1) : ( 3x + 1)

Vậy ( 3x⁴ + 8x³ – 2x² + x + 1) : ( 3x + 1) = x³ + $\frac{7}{3}$x² – $\frac{13}{9}$x + $\frac{22}{7}$ (dư $\frac{5}{27}$).