Question

giải pt:

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

2, ( x+2 ) ( x+4 ) ( x+6 ) (x+8 ) +16 = 0

3, ( x+2 ) ( x+3 ) (x+4 ) ( x+ 15 ) -24 = 0

 

Answer 1

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

\(\Leftrightarrow\frac{x}{\left(x+1\right)\left(x-1\right)}+\frac{3}{\left(x-3\right)\left(x+1\right)}=\frac{x}{\left(x-3\right)\left(x-1\right)}\)

\(\Leftrightarrow x\left(x-3\right)+3\left(x-1\right)=x\left(x+1\right)\)

\(\Leftrightarrow x^2-3=x^2+x\)

\(\Leftrightarrow-3=x\)

\(\Leftrightarrow x=-3\)

Vậy: nghiệm phương trình là -3

Answer 2

\(3,\text{ }\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=0\)

\(\Rightarrow\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=0-16\)

\(\Rightarrow\text{ Có lẻ thừa số âm }\)

Mà \(\left(x+8\right)>\left(x+6\right)>\left(x+4\right)>\left(x+2\right)\)

Ta có hai trường hợp : 

\(TH\text{ }1\text{ :}\) Có một thừa số âm

\(\Rightarrow\text{ }\left(x+2\right)< 0\)

\(\Rightarrow\text{ }x< -2\)

\(TH\text{ }2\text{ : }\) Có 3 thừa số âm

\(\Rightarrow\text{ }\hept{\begin{cases}\left(x+2\right)< 0\\\left(x+4\right)< 0\\\left(x+6\right)< 0\end{cases}}\)                \(\Rightarrow\text{ }\left(x+2\right)< 0\text{ }\Rightarrow\text{ }x< -2\)

Si thì thôi nha ! Mong bạn thông cảm !