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a)Ta có : \(4x^2=1\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào B , ta được:
\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)
Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)
b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)
\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)
\(=\frac{x}{x+1}\)
Vậy \(M=\frac{x}{x+1}\)
c)Ta có: \(x< x+1\forall x\)
\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)
Vậy với mọi \(x\ne-1\)thì \(M< 1\)
Do : \(4x^2=1\)
\(< =>\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta thấy điều kiện xác định của B là \(x\ne-\frac{1}{2}\)
Suy ra \(x=\frac{1}{2}\)
Ta có : \(B=\frac{x^2-x}{2x+1}=\frac{\frac{1}{4}-\frac{1}{2}}{\frac{1}{2}.2+1}=\frac{\frac{-1}{4}}{2}=-\frac{1}{8}\)
Vậy ......
Ta có : \(A=\frac{1}{x-1}+\frac{x}{x^2-1}=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{x^2-1}\)
Suy ra \(M=\frac{2x+1}{x^2-1}.\frac{x^2-x}{2x+1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x}{x+1}\)
Bài làm
a) \(M=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(M=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{1}{\left(x+1\right)\left(x-1\right)}\)
\(M=\left(\frac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right).\left(x+1\right)\left(x-1\right)\)
\(M=\frac{2x+1}{\left(x+1\right)\left(x-1\right)}.\left(x+1\right)\left(x-1\right)\)
\(M=2x+1\)
b) Thay x = 1/2 vào M ta được:
\(M=2.\frac{1}{2}+1\)
\(M=1+1=2\)
Vậy M = 2 khi x = 1/2.
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
\(ĐK:\hept{\begin{cases}x-1\ne0\\1+x\ne0\end{cases}\Rightarrow x\ne\pm1}\)
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
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.