Bài tập về đa thức cho đơn thức

1. 

a) $(21x^{2}y-7xy^{2}+xy):xy = 21x - 7y +1$

b) $(23x^{3}y^{2}-18x^{2}y^{3}+6xy^{5}):4xy^{2}=\frac{23}{4}x^{2}-\frac{9}{2}xy+\frac{3}{2}y^{3}$

c) $(5a^{3}b^{2}c+7a^{2}b^{3}c^{2}-4abc^{2}):a^{2}bc=5ab+7b^{2}c-4c$

d) $[6(x-2)^{3}+7(x-2)^{4}-9(x-2)^{5}]:(2-x)^{2}$

 = $[6(x-2)^{3}+7(x-2)^{4}-9(x-2)^{5}]:(x-2)^{2}$

 = $6(x-2)+7(x-2)^{2}-9(x-2)^{3}$

e) $[5(x-y)^{6}+2(x-y)^{5}-7(x-y)^{3}]:(y-x)^{2}$

 = $[5(x-y)^{6}+2(x-y)^{5}-7(x-y)^{3}]:(x-y)^{2}$

 = $5(x-y)^{4}+2(x-y)^{3}-7(x-y)$

2. Đặt $x^{2}+5$ = y ta có:

  $(x^{2}+6)(x^{2}+4):(x^{2}+5)$

= $(y+1)(y-1):y$

= $(y^{2}-1):y$

= $y-\frac{1}{y}$

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

3. 

a) $[7(x^{2}-1)^{4}+2(1-x)^{3}-3(x-1)^{2}]:2(x-1)^{2}$

 = $[7(x-1)^{4}(x+1)^{4}-2(x-1)^{3}-3(x-1)^{2}]:2(x-1)^{2}$

 = $\frac{7}{2}(x-1)^{2}(x+1)^{4}-(x-1)-\frac{3}{2}$

b) $[5(x^{3}-y^{3})^{4}+(x-y)^{3}]:(x^{2}-2xy+y^{2})$

 = $[5(x-y)^{4}(x^{2}+xy+y^{2})^{4}+(x-y)^{3}]:(x-y)^{2}$

 = $5(x-y)^{2}(x^{2}+xy+y^{2})^{4}+(x-y)$