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

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

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

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

.......... 

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

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

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

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

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

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

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

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

Vậy phương trình có nghiệm là : x = -2040

Link

bạn tham khảo link đó nhé

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

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

\(\ge2\sqrt{\left(\frac{2xy}{x^2+y^2}\right)^2.\left(\frac{x^2+y^2}{2xy}\right)^2}+3\left(\frac{2xy}{2xy}\right)^2-2=3\)

1.  x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)

2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)

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

áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)

x=1 nhe nhap minh di ma ket ban voi minh nhe

Bạn rút gọn sai rồi, mình nhìn đề bài b) cho x>2 thì là biết chắc bạn sai , mình làm lại nhé : ( ĐKXĐ : tự làm )

a) \(Q=\frac{x\left(x+2\right)}{\left(x-2\right)^2}:\left(\frac{\left(x+2\right)\left(x-2\right)+x+6-x^2}{x\left(x-2\right)}\right)\)

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

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

Vậy \(Q=\frac{x^2}{x-2}\)

b) Ta có : \(Q=\frac{x^2}{x-2}=\frac{x^2-4+4}{x-2}=x+2+\frac{4}{x-2}=x-2+\frac{4}{x-2}+4\)

Do \(x>2\Rightarrow x-2>0\) và \(\frac{4}{x-2}>0\)do đó áp dụng BĐT Cô si cho 2 số dương ta được :

\(x-2+\frac{4}{x-2}\ge2\sqrt{\left(x-2\right).\left(\frac{4}{x-2}\right)}=2\cdot\frac{1}{2}=1\)

\(\Rightarrow Q\ge1+4=5\)

Vậy : GTNN của \(Q=5\)

P/s : Ai vào kiểm tra hộ cái :)) Sợ sai lắm nhé, cảm ơn nha 33

Nếu chưa học Cô si thì chứng minh rồi dùng thôi :

Bài này sử dụng Cô - si hai số nên cần chứng minh BĐT :

\(a+b\ge2\sqrt{ab}\left(a,b>0\right)\)

Thật vậy : \(a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng )

Do đó \(a+b\ge2\sqrt{ab}\) với a,b >0

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

x² - 2x + 2013 / x² 

x² -2x + 1 + 2012 / x²

(x -1)² + 2012/x²

(x -1)²/x² +  2012/x²

^{GTNN là} 2012/x²  khi (x -1)² bàng 0 => x=1 ( khó viết :v)

Vì \(A=\frac{x^2-2x+2014}{\left(x+1\right)^2}\)

\(\Rightarrow x^2-2x+2014=A\left(x+1\right)^2\)

\(\Leftrightarrow x^2-2x+2014=Ax^2+2Ax+A\)

\(\Leftrightarrow\left(1-A\right)x^2-2\left(A+1\right)x+\left(2014-A\right)=0\)

\(\Delta=4\left(A+1\right)^2-4\left(1-A\right)\left(2014-A\right)\)

\(=8068A-8052\)

Vì A có GTNN nên phương trình có nghiệm

\(\Leftrightarrow8068A-8052\ge0\Leftrightarrow A\ge\frac{2013}{2017}\)

Dấu "=" khi \(x=\frac{2015}{2}\)