giải pt

R

riverflowsinyou1

12/8+x^3 = 1 + 1/x+2
DK: $x$ #2 , $x$ # $-2$
$\frac{12}{8+x^3}=1+\frac{1}{x+2}$
\Rightarrow $\frac{12.x+16-x^3}{(8+x^3)(x+2)}=1$
\Rightarrow $12x+16-x^3=(8+x^3)(x+2)=8.x+16+x^4+2.x^3$
\Rightarrow $12.x=8.x+x^4+3.x^3$
\Rightarrow $4.x=x^4+3.x^3$
\Leftrightarrow $x^4+3.x^3-4.x=x^4-x+3.x^3-3.x=x.(x^3-1)+3.x(x^2-1)=(x-1)(x^2-x+1).x+3.x.(x-1)(x+1)=(x-1)(x^3-x^2+x)+(x-1)(3.x^3+3x)=(x-1)(x^3-x^2+x+3.x^3+3.x)=(x-1)(4.x^3+4.x-x^2)=0$
TH1: $x-1=0$ \Rightarrow $x=1$
TH2: $4.x^3+4.x-x^2=x.(4.x^2+4-x)=0$
$x=0$ hay $4.x^2-x+4=x^2-0,25.x+1=0$ \Rightarrow vô nghiệm
\Rightarrow $x$ $\in$ ${0;1}$
 
H

hocmai.toanhoc

Bài này nên lấy $x^3+8=(x+2)(x^2-2x+4)$ làm mẫu thức chung sẽ ngắn và gọn hơn rất nhiều
 
R

riverflowsinyou1

(x^3-x^2-x+1) = 3/(1-x^2) - 1/x+1
$(x^2-1)(x-1)=\frac{3}{(1-x)(x+1)}-\frac{1}{x+1}$
$(x^2-1)(x-1)=\frac{1}{x+1}.($\frac{3}{1-x}$-1}$
$(x^2-1)^2=\frac{3}{1-x}-1$
$x^4-2.x^2+2=\frac{3}{1-x}$
$(1-x)(x^4-2.x^2+2)=-x^5+x^4+2.x^3-2.x^2-x2+2=3$
$x^5+x^4+2.x^3-2.x^2-2.x-1=0$ chừ giải pt thôi :)
 
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