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 \(M=\frac{2x+1}{x^2-1}.\frac{x^2-x}{2x+1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x}{x+1}\)
a.Ta có:\(4x^2=1\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\left(tmđkxđ\right)\)
Với \(x=\frac{1}{2}\left(tm\right)\)thì biểu thức có giá trị là:
\(\frac{x^2-x}{2x+1}=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=-\frac{1}{8}\)
b.Ta có:\(M=A.B\)
\(=\left(\frac{1}{x-1}-\frac{x}{1-x^2}\right).\frac{x^2-x}{2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right).\frac{x^2-x}{2x-1}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right).\frac{x^2-x}{2x+1}\)
\(=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x-1}=\frac{\left(2x+1\right)x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2x+1\right)}\)
\(=\frac{x}{x+1}\)
c.Để \(M< 1\)thì \(\frac{x}{x+1}< 1\)
Tự giải bất phương trình nha :))
