Question

Bài 1:Tìm đa thức M

a)\(\dfrac{^{x^3}+27}{x^2-3x+9}\)=\(\dfrac{x+3}{M}\)

b)\(\dfrac{M}{x+4}\)=\(\dfrac{x^2-8x+16}{16-x^2}\)

c)\(\dfrac{x-2y}{M}\)=\(\dfrac{3x^2-7xy+2y^2}{3x^2+5xy-2y^2}\)

Answer 1

Answer 2

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

Answer 3

Answer 4

a) Ta có: \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Leftrightarrow M=\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{\left(x+3\right)\left(x^2-3x+9\right)}=1\)

b) Ta có: \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}\)

\(\Leftrightarrow M=\dfrac{\left(x-4\right)^2\cdot\left(x+4\right)}{\left(4-x\right)\left(4+x\right)}=4-x\)