Question

[umi]

Cho biểu thức M=\((\)\(\frac{1}{x-1}\)-    \(\frac{x}{1-x^3}\)   x   \(\frac{x^2+x+1}{x+1}\)\()\) : \(\frac{1}{x^2-1}\)

a) rut gon M

b) tính giá trị của M khi x = 1/2

c) tìm giá trị của x để M luôn co gia tri dương

Answer 1

[Nguyệt]

\(ĐK:\hept{\begin{cases}x-1\ne0\\1+x\ne0\end{cases}\Rightarrow x\ne\pm1}\)

a) \(M=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right).\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x+1}\right)\cdot\frac{x^2-1}{1}\)

\(M=\left(\frac{1}{x-1}-\frac{x}{1-x}\right)\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}\)

\(M=\left(\frac{x+1}{\left(x-1\right).\left(x+1\right)}-\frac{-x}{\left(x-1\right).\left(x+1\right)}\right)\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}\)

\(M=\frac{2x+1}{\left(x-1\right).\left(x+1\right)}\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}=2x+1\)

b) \(M=2x+1=\frac{2.1}{2}+1=1+1=2\)

c) \(M=2x+1>0\Rightarrow2x>-1\Rightarrow x>-\frac{1}{2}\)và x khác +1,-1

Answer 2

[Huy Hoàng]

a/ Ta có \(M=\frac{\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}}{\frac{1}{x^2-1}}\) với \(x\ne\pm1\)

\(M=\frac{\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}}{\frac{1}{\left(x-1\right)\left(x+1\right)}}\)

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

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

\(M=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}.\left(x-1\right)\left(x+1\right)\)

\(M=2x+1\)

b/ Ta có \(x=\frac{1}{2}\)thoả mãn ĐKXĐ

Vậy với \(x=\frac{1}{2}\):

\(M=2x+1=2.\frac{1}{2}+1=2\)

c/ Khi M > 0

=> \(2x+1>0\)

=> \(x>-\frac{1}{2}\)

Vậy khi \(\hept{\begin{cases}x>-\frac{1}{2}\\x\ne\pm1\end{cases}}\)thì M > 0.