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 \(M< 1\)

a) \(ĐKXĐ:x\ne\pm1\)

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

\(\Leftrightarrow Q=\frac{1}{2\left(x-1\right)}+\frac{1}{2\left(x+1\right)}-\frac{x^2}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow Q=\frac{x+1+x-1-2x^2}{2\left(x+1\right)\left(x-1\right)}\)

\(\Leftrightarrow Q=\frac{-2x^2+2x}{2\left(x+1\right)\left(x-1\right)}\)

\(\Leftrightarrow Q=\frac{-1}{x+1}\)

b) Khi \(\left|x+1\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-3\left(tm\right)\end{cases}}\)

Thay \(x=-3\)vào Q ta được :

 \(Q=\frac{-1}{-3+1}=\frac{1}{2}\)

c) Để \(Q\)có giá trị nguyên \(\Leftrightarrow-1⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(-1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow x\in\left\{-2;0\right\}\)

Vậy để Q có giá trị nguyên \(\Leftrightarrow x\in\left\{-2;0\right\}\)

c) Bạn lấy mỗi giá trị nguyên nhỏ nhất của x = -2 thôi nhé !

Xin lỗi vì đọc nhầm đề

a, Ta có : \(A=\frac{1}{x+2}-\frac{2x}{4-x^2}+\frac{3}{x-2}\)

\(=\frac{1}{x+2}-\frac{2x}{\left(2-x\right)\left(x+2\right)}+\frac{3}{x-2}\)

\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra : \(M=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{3x+2}\)

\(=\frac{2\left(3x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}=\frac{2}{x-2}\)

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

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

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

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

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 :))