Cho m+n=1 , m.n khác 0 , chứng minh :
\(\frac{m}{n^3-1} \)+ \(\frac{n}{m^3-1}\)= \(\frac{2\left(mn-2\right)}{m^2n^2+3}\)
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YN
25 tháng 11 2021
Answer:
a) \(\frac{x}{x-1}+\frac{3}{x+1}-\frac{6x-4}{x^2-1}\) \(\left(ĐKXĐ:x\ne\hept{\begin{cases}x=1\\x=-1\end{cases}}\right)\)
\(=\frac{x}{x-1}+\frac{3}{x+1}-\frac{6x-4}{\left(x+1\right).(x-1)}\)
Mẫu thức chung: \(\left(x+1\right).\left(x-1\right)\)
\(B=\frac{x.\left(x+1\right)}{\left(x+1\right).\left(x-1\right)}+\frac{3.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{6x-4}{\left(x+1\right).\left(x-1\right)}\)
\(=\frac{x}{\left(x+1\right).\left(x-1\right)}+\frac{3x-3}{\left(x+1\right).\left(x-1\right)}-\frac{6x-4}{\left(x+1\right).\left(x-1\right)}\)
\(=\frac{x^2+x+3x-3-6x+4}{\left(x+1\right).\left(x-1\right)}\)
\(=\frac{x^2-2x+1}{\left(x+1\right).\left(x-1\right)}\)
\(=\frac{x-1}{x+1}\)
b) \(B=5\)
\(\Rightarrow\frac{x-1}{x+1}=5\)
\(\Rightarrow x-1=5.\left(x+1\right)\)
\(\Rightarrow x-1=5x+5\)
\(\Rightarrow x-1-5x-5=0\)
\(\Rightarrow-4x-6=0\)
\(\Rightarrow-2.\left(2x+3\right)=0\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=\frac{-3}{2}\)
YN
25 tháng 11 2021
Answer:
a) ĐKXĐ: \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)
\(A=\left(\frac{1}{x}+\frac{x}{x+1}\right):\frac{x}{x^2+x}\)
\(=\frac{x+1+x.x}{x.\left(x+1\right)}.\frac{x.\left(x+1\right)}{x}\)
\(=\frac{x^2+x+1}{x}\)
b) \(x=4\)
\(\Rightarrow A=\frac{4^2+4+1}{4}=\frac{21}{4}\)
c) \(A=3\)
\(\Rightarrow\frac{x^2+x+1}{x}=3\)
\(\Rightarrow x^2+x+1=3x\)
\(\Rightarrow x^2-2x+1=0\)
\(\Rightarrow\left(x-1\right)^2=0\)
\(\Rightarrow x=1\left(TMĐK\right)\)