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a) A = ( \(\frac{x+1}{x-1}\)\(-\)\(\frac{x-1}{x+1}\)) \(\div\)\(\frac{2x}{5x-5}\)
= ( \(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\)\(-\)\(\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)) \(\div\)\(\frac{2x}{5x-5}\)
= \(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)\(\div\)\(\frac{2x}{5x-5}\)
= \(\frac{\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x-1\right)\left(x+1\right)}\)\(\times\)\(\frac{5\left(x-1\right)}{2x}\)
= \(\frac{4x}{\left(x-1\right)\left(x+1\right)}\)\(\times\)\(\frac{5\left(x-1\right)}{2x}\)
= \(\frac{10}{x+1}\)
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 ??
2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)
Khi |x - 1| = 2
=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
Khi x = - 1 (không thỏa mãn) => Không tìm được A
b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
Đẻ P < 8
=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)
=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)
Vậy x > - 1 thì P < 8
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, \(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
b, Thay x = -2 ta được :
\(\frac{x+1}{x-1}=\frac{-2+1}{-2-1}=\frac{1}{3}\)
Vậy A nhận giá trị 1/3
\(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right)\div\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{x-1}\)
Với x = -2 (tmđk) => \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)
1, \(=\left[\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right]:\frac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(1+x+x^2-x\right):\frac{1-x^2}{\left(1-x\right)\left(1-x^2\right)}\)\(=\left(x^2+1\right)\left(1-x\right)\)
2, để B<0 <=> (x2+1)(1-x)<0
vì x^2+1 > 0 với mọi x
=> \(\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}\Leftrightarrow x>1}\)
3, \(\left|x-4\right|=5\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\left(loại\right)\end{cases}}\)
Thay x=9 vào B ta có: B=(92+1)(1-9)=82.(-8)=-656
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 \(M=\frac{2x+1}{x^2-1}.\frac{x^2-x}{2x+1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x}{x+1}\)