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

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

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

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

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

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

\(B=\frac{x}{x+1}\)

MÌnh nghĩ đề câu b là với x>-4 mới đúng chứ

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

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

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

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

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

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

Áp dụng BĐT Cô - Si cho 2 số dương \(x-2;\frac{4}{x-2}\)ta có :

\(x-2+\frac{4}{x-2}\ge2\sqrt{\frac{\left(x-2\right).4}{x-2}}=2\sqrt{4}=4\)

\(\Rightarrow x-2+\frac{4}{x-2}\ge4\Rightarrow x-2+\frac{4}{x-2}+4\ge8\)

Hay \(S_{min}=4\Leftrightarrow x-2=\frac{4}{x-2}\)

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

\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)\(\Rightarrow...\)

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{8x}{x^2-1}\right):\left(\frac{2x-2x^2-6}{x^2-1}-\frac{2}{x-1}\right)\)

\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{8x}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{2x-2x^2-6}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)

\(A=\left(\frac{x^2+2x+1-x^2+2x-1-8x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x-2x^2-6-2x-2}{\left(x+1\right)\left(x-1\right)}\right)\)

\(A=\left(\frac{4x-8x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{-2x^2-8}\)

.......... 

\(\frac{x+32}{2008}+\frac{x+31}{2009}+\frac{x+29}{2011}+\frac{x+28}{2012}+\frac{x+2056}{4}=0\) \(=0\)

\(\Leftrightarrow\)\(\frac{x+32}{2008}+1+\frac{x+31}{2009}+1+\frac{x+29}{2011}+1\)\(+\frac{x+28}{2012}+1+\frac{x+2056}{4}-4\)\(=0\)

\(\Leftrightarrow\)\(\frac{x+32}{2008}+\frac{2008}{2008}+\frac{x+31}{2009}+\frac{2009}{2009}+\)\(\frac{x+29}{2011}+\frac{2011}{2011}+\frac{x+28}{2012}+\frac{2012}{2012}+\)\(\frac{x+2056}{4}-\frac{16}{4}\)\(=0\)

\(\Leftrightarrow\)\(\frac{x+32+2008}{2008}+\frac{x+31+2009}{2009}\)\(+\frac{x+29+2011}{2011}+\frac{x+28+2012}{2012}\)\(+\frac{x+2056-16}{4}\)\(=0\)

\(\Leftrightarrow\)\(\frac{x+2040}{2008}+\frac{x+2040}{2009}+\frac{x+2040}{2011}\)\(+\frac{x+2040}{2012}+\frac{x+2040}{4}=0\)

\(\Leftrightarrow\)\(\left(x+2040\right).\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+2040=0\\\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}=0\end{cases}}\)(vô lí)

\(\Leftrightarrow\)\(x=-2040\)

Vậy phương trình có nghiệm là : x = -2040

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 \(x=0\)

a/ ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)

\(A=\left[\frac{1}{\sqrt{x}-1}+\frac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}+4}{\sqrt{x}-1}\right]\)

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

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

b/ 

Ta có: \(A=\frac{1}{\sqrt{x}+1}\ge1\)

                                          Vậy Min A = 1 .Dấu "=" xảy ra khi x = 0

a , rút gọn : A= \(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{x-1}\right):\left(2-\frac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)

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

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

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

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

                   A = \(\frac{1}{\sqrt{x}+1}\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2