a. tìm điều kiện xác định của P

ĐKXĐ: \(x\ne0;x\ne\pm1\)

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

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

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

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

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

\(P=\frac{x}{x-1}\)

b. tìm x 

Với P = 2 ta có:

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

=>  x = 2(x-1)

=> x = 2x -2

=> 2x - x = 2

=> x = 2

Vậy với x = 2 thì P = 2

c. với 0 < x < 1 . hãy so sánh P với |P|

\(P=\frac{x}{x-1}\)

Với 0< x < 1 thì x -1 <0 ; x>0 => P <0 

Suy ra P< |P| ( vì |P| >0)

A. DE P XAC DINH

<=>X^2-1 KHÁC 0<=>X KHAC -1 VÀ X KHÁC 1

<=>2X+2 KHAC 0 <=>X KHAC-1

<=>2X KHAC 0 <=>X KHAC 0

=> X KHAC O HOAC X KHAC +-1

TACO:( 2X / X^2-1 +X-1/ 2X+2 ) : X+1 / 2X

=[2X . 2 / (X+1)(X-1). 2  + (X-1)(X-1) / 2(X+1)(X-1) ] : X+1/2X

=[4X+(X-1)^2]  /  2(X+1)(X-1)  :X+1 / 2X

=(4X+X^2-2X+1) / 2(X+1)(X-1)  : X+1/2X

=X^2+2X+1 / 2(X-1)(X+1) : X+1 / 2X

=(X+1)^2 / 2(X-1)(X+1) : X+1/2X

=(X+1) / 2(X-1) . 2X/X+1

=X/X-1

B. DE P=2

<=>X/X-1=2

<=>X=2(X-1)=2X-2=X+X-2

TA CÓ: X +X-2 = X+0

=>X-2=0

=>X=2

C .VI 0<X<1

=>X / X-1 = |X/X-1|

=>P=|P|

Câu 1 :

a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)

\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)

\(\Leftrightarrow2x^2+8x+6=0\)

\(\Leftrightarrow x^2+4x+4-1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)

Vậy : \(x=-3\) thì P = 1.

1111111

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

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

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

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

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

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

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

\(M=\frac{0+x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

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

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