Giải SBT Toán 7 CD Bài 5. Phép chia đa thức một biến có đáp án

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

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

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

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

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

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

= –3x$^{2}$ + 9x – 18;

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

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

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

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

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

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

a) $P(x)=(6{x}^{5}-\frac{1}{2}{x}^{4}+\frac{1}{3}{x}^{3}):(2{x}^{3})$

$=(6{x}^{5}:2{x}^{3})-(\frac{1}{2}{x}^{4}:2{x}^{3})+(\frac{1}{3}{x}^{3}:2{x}^{3})$

$=(6:2).({x}^{5}:{x}^{3})-(\frac{1}{2}:2).({x}^{4}:{x}^{3})+(\frac{1}{3}:2).({x}^{3}:{x}^{3})$

$=3.{x}^{5-3}-\frac{1}{4}.{x}^{4-3}+\frac{1}{6}.{x}^{3-3}$

$=3.{x}^{2}-\frac{1}{4}.{x}^{1}+\frac{1}{6}.{x}^{0}$

$=3{x}^{2}-\frac{1}{4}x+\frac{1}{6}.$

Thay x = –2 vào $P(x)=3{x}^{2}-\frac{1}{4}x+\frac{1}{6}$ ta được:

$P(-2)=3.{(-2)}^{2}-\frac{1}{4}.(-2)+\frac{1}{6}=3.4+\frac{1}{2}+\frac{1}{6}$

$=\frac{72+3+1}{6}=\frac{76}{6}=\frac{38}{3}.$

Vậy tại x = –2 thì P(x) có giá trị bằng $\frac{38}{3}$