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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.
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|
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). 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, P= \(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\): ( \(\frac{x+1}{x}\)+ \(\frac{1}{x-1}\)- \(\frac{x^2-2}{x\left(x-1\right)}\)
P= \(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\): \(\frac{\left(x+1\right)\left(x-1\right)+x-x^2+2}{x\left(x-1\right)}\)
P= \(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\). \(\frac{x\left(x-1\right)}{x^2-1+x-x^2+2}\)
P= \(\frac{x^2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
P= \(\frac{x^2}{x-1}\)( đkxđ x khác 1)
b, để P=\(\frac{-1}{2}\)\(\Rightarrow\)\(\frac{x^2}{x-1}\)=\(\frac{-1}{2}\)\(\Rightarrow\)1-x = 2x\(^2\)
\(\Rightarrow\)2x\(^2\)+ x-1 = 0\(\Rightarrow\)2x\(^2\)- 2x +x - 1 =0\(\Rightarrow\)(x -1 ) (2x + 1) = 0
\(\Rightarrow\)\(\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\orbr{\begin{cases}x=1\left(ktm\right)\\x=\frac{-1}{2}\left(tm\right)\end{cases}}\)
vậy x= \(\frac{-1}{2}\)
c, tớ chịu thôi mà tớ mỏi tay lắm òi. k cho tớ nhé
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)
Rút gọn :
\(P=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{\left(x+1\right)\left(x-1\right)+x+\left(2-x^2\right)\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x^2-1+x+2x-2-x^3+x^2}{x\left(x-1\right)}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{-x^3+2x^2+3x-3}{x\left(x-1\right)}\right)\)
a)
2x-4=2(x-2)
2x+4=2(x+2)
x
Để P xác định thì
[2(x-2) => [2(x+2)
[2(x+2) =>[ 2(x-2)
[ (x-2)(x+2) => [(x+2)(x-2)
Vay 2(x+2) , 2(x-2), (x+2)(x-2) thi P xác định
a) ĐKXĐ: x khác +-1
b) \(\frac{x+1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{x^2-1}\)
\(=\frac{x+1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2+x+5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2+\left(x-2\right)\left(x-1\right)-\left(2x^2+x+5\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=-\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=-\frac{2}{x-1}\)