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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 \(M=\frac{2x+1}{x^2-1}.\frac{x^2-x}{2x+1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x}{x+1}\)
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 : \(M=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{3x+2}\)
\(=\frac{2\left(3x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}=\frac{2}{x-2}\)
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). dấu = xảy ra khi x2=0
=> x=0
Vậy \(B_{min}\Leftrightarrow x=0\)
ta có: \(x^2+2x-2=x^2+2x+1^2-3=\left(x+1\right)^2-3\ge-3\)
dấu = xảy ra khi \(x+1=0\)
\(\Rightarrow x=-1\)
Vậy\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)
a) \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)(với \(x\ne\pm2;x\ne-1\))
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{-\left(6-5x\right)}{x^2-4}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{1}{x-2}:\frac{x+1}{x-2}=\frac{1}{x-2}\cdot\frac{x-2}{x+1}=\frac{1}{x+1}\)
b) Với \(M=\frac{1}{4}\)ta có :
\(M=\frac{1}{x+1}\Rightarrow\frac{1}{4}=\frac{1}{x+1}\)
\(\Rightarrow1\left(x+1\right)=4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a, \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}=\frac{1}{x-2}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : M = 1/4 hay \(\frac{1}{x+1}=\frac{1}{4}\Leftrightarrow4=x+1\Leftrightarrow x=3\)
a) M= \(\frac{2x}{1-x^2}\)( \(\frac{1}{x^2+2x+1}\)- \(\frac{1}{x^2-1}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1}{x^2+2x+1}\)+ \(\frac{1}{1-x^2}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1}{\left(x+1\right)^2}\)+ \(\frac{1}{\left(1+x\right)\left(1-x\right)}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1-x}{\left(1-x\right)\cdot\left(x+1\right)^2}\)+ \(\frac{1+x}{\left(1-x\right)\cdot\left(x+1\right)^2}\))
= \(\frac{2x}{1-x^2}\)(\(\frac{1-x^2}{\left(1-x\right)\cdot\left(x+1\right)^2}\))
= \(\frac{2x}{\left(1-x\right)\cdot\left(x+1\right)^2}\)