Giải SBT Toán 7 CD Bài 3. Phép cộng, phép trừ đa thức một biến có đáp án

a) Ta có: F(x) + Q(x) = x$^{5}$ – x$^{3}$ + 2.

Suy ra Q(x) = x$^{5}$ – x$^{3}$ + 2 – F(x)

Hay Q(x) = x$^{5}$ – x$^{3}$ + 2 – (x$^{7}$ – $\frac{1}{2}$x$^{3}$ + x + 1)

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

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

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

Vậy Q(x) = – x$^{7}$ + x$^{5}$ – $\frac{1}{2}$x$^{3}$ – x + 1.

b) Ta có: F(x) – R(x) = 2.

Suy ra R(x) = F(x) – 2.

Hay R(x) = x$^{7}$ – $\frac{1}{2}$x$^{3}$ + x + 1 – 2.

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

Vậy R(x) = x$^{7}$ – $\frac{1}{2}$x$^{3}$ + x – 1.

Ta có P(x) + Q(x) = x$^{2}$ + 1 và P(x) – Q(x) = 2x.

Suy ra [P(x) + Q(x)] + [P(x) – Q(x)] = (x$^{2}$ + 1) + 2x.

Hay 2P(x) = x$^{2}$ + 2x + 1.

Do đó P(x) = $\frac{1}{2}$x$^{2}$ + x + $\frac{1}{2}$.

Mặt khác: P(x) – Q(x) = 2x

Suy ra Q(x) = P(x) – 2x

Hay Q(x) = $\frac{1}{2}$x$^{2}$ + x + $\frac{1}{2}$ – 2x

= $\frac{1}{2}$x$^{2}$ – x + $\frac{1}{2}$.

Vậy P(x) = $\frac{1}{2}$x$^{2}$ + x + $\frac{1}{2}$ và Q(x) = $\frac{1}{2}$x$^{2}$ – x + $\frac{1}{2}$.

a) Ta có:

H(x) = F(x) + G(x).

= (x$^{4}$ + x$^{3}$ – 3x$^{2}$ + 2x – 9) + (– x$^{4}$ + 2x$^{2}$ – x + 8)

= x$^{4}$ + x$^{3}$ – 3x$^{2}$ + 2x – 9 – x$^{4}$ + 2x$^{2}$ – x + 8

= (x$^{4}$ – x$^{4}$) + x$^{3}$ + (– 3x$^{2}$ + 2x$^{2}$) + (2x – x) + (– 9 + 8)

= x$^{3}$ – x$^{2}$ + x – 1.

Vậy H(x) = x$^{3}$ – x$^{2}$ + x – 1.