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

Đk:\(x\ne0;1;2;3;4\)

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

\(\Leftrightarrow\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x}=2-\frac{1}{4-x}\)

\(\Leftrightarrow\frac{1}{x-4}-\frac{1}{x}=2-\frac{1}{4-x}\)\(\Leftrightarrow\frac{4}{x\left(x-4\right)}=\frac{2x-7}{x-4}\)

Dễ thấy \(x\ne4\) nên nhân 2 vế của pt vừa biến đổi với \(x-4\) ta dc:

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

\(\Leftrightarrow2x^2-7x=4\Leftrightarrow2x^2-7x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+1\right)=0\)\(\Leftrightarrow x=-\frac{1}{2}\left(x\ne4\right)\)

a, tự lm......

P=x² / x-1

b, P<1

=> x²/x-1  <1

<=>x²/x-1 -1 <0

<=>x²-x+1 / x-1<0

Vi x²-x+1= (x -1/2 )²+3/4 >0

=> Để P<1

x-1 <0

x <1

c, x²/x-1 = x²-1+1/x-1

             = x+1 +1/x-1

               = 2 +(x-1) + 1/x-1

Áp dụng BDT Cô si ta có :

x-1  + 1/x-1 >hoặc = 2

=> P>= 3

Đầu = xảy ra <=> x=2( x >1)

Vay......

làm đúng nhuwng phần c, phải >=4 cơ vì công cả 2 vế với 2 ta có P>=4

1/ Ta có : \(\frac{\left(x+2\right)+\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\frac{1}{x-2}\)

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

=>  \(\left(2x+1\right)\left(x-2\right)=\left(x-1\right)\left(x+2\right)\)

=>   \(2x^2-3x-2=x^2+x-2\)

=>    \(x^2-4x=0\)

=>    \(x\left(x-4\right)=0\)

=>    \(\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

2/ Ta có:   \(\frac{x+3+2\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}=\frac{3}{x+2}\)

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

=>    \(\frac{3x+5}{\left(x+1\right)\left(x+3\right)}=\frac{3}{x+2}\)

=>    \(\left(x+1\right)\left(x+3\right).3=\left(3x+5\right)\left(x+2\right)\)

=>     \(3x^2+12x+9=3x^2+11x+10\)

=>     \(x=1\)

Kết quả là 25

1)   \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)

\(\Leftrightarrow\)   \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1

\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)+ \(\frac{x +2009}{2006}\)= \(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)

\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)

\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0

Ta thấy:  1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0

\(\Leftrightarrow\)x + 2009 = 0

\(\Leftrightarrow\)x = -2009

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

tìm đc x=0;1;2