rút gọn phần phân thức đại số

N

nhimcon_online

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T

trinhminh18

a/Ta có:
$\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}$=$\dfrac{(1-x)(3y-2)}{(1-x)(x^2+x+1-3x)}$
=$\dfrac{3y-2}{(x-1)^2}$
 
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trinhminh18

b)A= $\dfrac{x^{40} +x^{30} +x^{20}+x^{10} +1}{x^{45} + x^{40} +x^{35} + ...+ x^{10} +x^5 +1}$
\Rightarrow$A.x^5=\dfrac{x^{45} +x^{35} +x^{25}+x^{15} +x^5}{x^{45} + x^{40} +x^{35} + ...+ x^{10} +x^5 +1}$
\Rightarrow$A.x^5+A=\dfrac{x^{45} +x^40+x^{35} +x^{25}+x^{15} +x^5+x^{40} +x^{30} +x^{20}+x^{10} +1}{x^{45} + x^{40} +x^{35} + ...+ x^{10} +x^5 +1}$
Hay $A(x^5+1)=1$
\Rightarrow$A=\dfrac{1}{x^5+1}$
 
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