\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)

\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)

\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)

\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)

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

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

b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)

thay \(x=-\frac{3}{2}\)  vào P tâ đc:   \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)

c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)

\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)

\(\Leftrightarrow4x+2=x^2-3x\)

\(\Leftrightarrow x^2-7x-2=0\)

\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)

bạn tự giải nốt nhé!!

d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)

\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

bạn tự làm nốt nhé

a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)

\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)

b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)

\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)

\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)

\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)

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

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

\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(KOTM\right)\\x=-\frac{3}{2}\left(TMđkcđ\right)\end{cases}}}\)

thay x=-3/2 vào P ta đc:\(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}+3}=\frac{4}{3}\)

c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x+3}=\frac{x}{2}\)

\(\Rightarrow2.\left(2x+1\right)=x.\left(x+3\right)\)

\(\Leftrightarrow4x+2=x^2+3x\)

\(\Leftrightarrow x^2-x-2=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}\left(TMđkxđ\right)}\)

d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x+3}\in Z\Leftrightarrow\frac{2x+6-5}{x+3}=2-\frac{5}{x+3}\in Z\)

\(2\in Z\Rightarrow\frac{5}{x+3}\in Z\Leftrightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

tự giải nốt

\(M=\frac{4x+8}{x^2-1}:\frac{x+2}{x+1}-\frac{x-2}{1-x}\)   \(ĐKXĐ:x\ne\pm1\)

\(M=\frac{4\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+1}{x+2}+\frac{x-2}{x-1}\)

\(M=\frac{4}{x-1}+\frac{x-2}{x-1}\)

\(M=\frac{4+x-2}{x-1}\)

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

vậy \(M=\frac{x+2}{x-1}\)

a

\(ĐKXĐ:x\in R\)

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

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

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

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

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

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

b

Xét \(x>0\Rightarrow M>0\)

Xét \(x=0\Rightarrow M=0\)

Xét \(x< 0\Rightarrow M>0\)

Vậy \(M_{min}=0\) tại \(x=0\)

\(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+4x^2+3}\left(ĐKXĐ:x\in R\right)\).

\(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+3\right)\left(x^2+1\right)}\).

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

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

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

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

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

Vậy với \(x\in R\)thì \(M=\frac{x^2}{x^4-x^2+1}\).

b) \(M=\frac{x^2}{x^4-x^2+1}\left(x\in R\right)\).

\(\Rightarrow\frac{1}{M}=\frac{x^4-x^2+1}{x^2}=x^2-1+\frac{1}{x^2}\).

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

\(\Rightarrow\frac{1}{M}=\left(x-\frac{1}{x}\right)^2+2-1=\left(x-\frac{1}{x}\right)^2+1\).

Ta có:

\(\left(x-\frac{1}{x}\right)^2\ge0\forall x\).

\(\Leftrightarrow\left(1-\frac{1}{x}\right)^2+1\ge1\forall x\).

\(\Leftrightarrow\frac{1}{M}\ge1\forall x\).

\(\Rightarrow M\le1\forall x\).

Dấu bằng xảy ra.

\(\Leftrightarrow x-\frac{1}{x}=0\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\).

Vậy \(maxM=1\Leftrightarrow x=\pm1\).