Giải SBT Toán 10 Bài 3. Nhị thức Newton có đáp án

a) Theo công thức nhị thức Newton ta có:

= x$^{4}$ + 4.x$^{3}$.3y + 6.x$^{2}$.(3y)$^{2}$ + 4.x.(3y)$^{3}$ + (3y)$^{4}$

= x$^{4}$ + 12x$^{3}$y + 54x$^{2}$y$^{2}$ + 108xy$^{3}$ + 81y$^{4}$.

b) Theo công thức nhị thức Newton ta có:

= 3$^{5}$ + 5.3$^{4}$.(–2x) + 10.3$^{3}$.(–2x)$^{2}$ + 10.3$^{2}$. (–2x)$^{3}$ + 5.3. (–2x)$^{4}$ + (–2x)$^{5}$

= 243 – 810x + 1080x$^{2}$ – 720x$^{3}$ + 240x$^{4}$ – 32x$^{5}$.

c) Theo công thức nhị thức Newton ta có:

${(x-\frac{2}{x})}^{5}$= x$^{5}$ + 5.x$^{4}$. $(\frac{-2}{\text{x}})$ + 10. x$^{3}$. ${(\frac{-2}{\text{x}})}^{2}$ + 10. x$^{2}$. ${(\frac{-2}{\text{x}})}^{3}$+ 5. x. ${(\frac{-2}{\text{x}})}^{4}$+${(\frac{-2}{\text{x}})}^{5}$

= x$^{5}$ – 10x$^{3}$ + 40x – $\frac{\text{80}}{\text{x}}$ + $\frac{\text{80}}{{\text{x}}^{3}}$– $\frac{32}{{\text{x}}^{5}}$ .

d) Theo công thức nhị thức Newton ta có:

${(3\sqrt{x}-\frac{1}{\sqrt{x}})}^{4}$= (3 $\sqrt{\text{x}}$)$^{4}$ + 4. (3 $\sqrt{\text{x}}$)$^{3}$. $(\frac{-1}{\sqrt{\text{x}}})$ + 6. (3 )$^{2}$. ${(\frac{-1}{\sqrt{\text{x}}})}^{2}$ + 4. (3 ).${(\frac{-1}{\sqrt{\text{x}}})}^{3}$ + ${(\frac{-1}{\sqrt{\text{x}}})}^{4}$ = 81x$^{2}$ – 108x + 54 – $\frac{\text{12}}{\text{x}}$+$\frac{\text{1}}{{\text{x}}^{\text{2}}}$

Ta có:

(2x + 1)$^{4}$ = (2x)$^{4}$ + 4.(2x)$^{3}$.1 + 6.(2x)$^{2}$.1$^{2}$ + 4.2x.1$^{3}$ + 1$^{4}$

= 16x$^{4}$ + 32x$^{3}$ + 24x$^{2}$ + 8x + 1

Ta có

$(x-2){(2x+1)}^{4}$

= ( x – 2 ) . ( 16x$^{4}$ + 32x$^{3}$ + 24x$^{2}$ + 8x + 1)

= 16x$^{5}$ + 32x$^{4}$ + 24x$^{3}$ + 8x$^{2}$ + x – 32x$^{4}$ – 64x$^{3}$ – 48x$^{2}$ – 16x – 2

= 16x$^{5}$ – 40x$^{3}$ – 40x$^{2}$ – 15x – 2.