\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010+18}{6}=0\)

\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010}{6}+3=0\)

\(\Rightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2010}{6}\right)=0\)

\(\Rightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+6\right)=0\)

Vì :\(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)

=> x + 2010 = 0

=> x = -2010

b) \(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)

\(\Rightarrow\frac{x-3}{2011}+\frac{x-2}{2012}-\frac{x-2012}{2}-\frac{x-2011}{3}=0\)

\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-2012}{2}-1\right)-\left(\frac{x-2011}{3}-1\right)=0\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)

\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)=0\)

Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\)

=> x - 2014 = 0

=> x = 2014

c) \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)

\(\Rightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)

\(\Rightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)

\(\Rightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)

Vì :\(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)

=> x + 66 = 0

=> x = -66

1) 1

2)Ta có: 2011 x 2013 + 2012 x 2014 =8100311

2012² + 2013² - 2 =8100311 . 

Vậy ta đã thấy 2 số bằng nhau

Kết luận : 2011 x 2013 + 2012 x 2014 = 2012²+ 2013² - 2

1, \(B=3^{24}-\left(27^4+1\right)\left(9^6-1\right)\)

\(=\left(3^{12}\right)^2-\left(3^{12}+1\right)\left(3^{13}-1\right)\)

\(=\left(3^{12}\right)^2-\left[\left(3^{12}\right)^2-1\right]\)

\(=\left(3^{12}\right)^2-\left(3^{12}\right)^2+1\)

\(=1\)

Vậy \(B=1\)

1) 1

2) hai số bằng nhau

1.

<=> B=\(3^{24}-\left\lbrack\left(3^3\right)^4+1\right\rbrack\left\lbrack\left(3^2\right)^6-1\right\rbrack\)

\(B=3^{24}-\left(3^{12}+1\right)\left(3^{12}-1\right)\)

\(B=3^{24}-3^{24}+1\)

\(B=1\)

2.

xét vế đầu tiên

\(2011\cdot2013+2012\cdot2014\)

\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)

\(=2012^2-1+2013^2-1\)

\(=2012^2+2013^2-2\)

=> \(2011\cdot2013+2012\cdot2014=2012^2+2013^2-2\)

a) \(=-7\left(x^2-\frac{10}{7}x+\frac{2016}{7}\right)\)

      \(=-7\left(x^2-2.\frac{5}{7}x+\frac{25}{49}+\frac{14087}{49}\right)\)

       \(=-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\)

ta có

\(\left(x-\frac{5}{7}\right)^2\ge0\)với mọi x

\(=>-7\left(x-\frac{5}{7}\right)^2\le0\)(nhân cả hai vế với -7)

\(=>-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\le-\frac{14087}{7}\)

trường hợp dấu "=" xảy ra khi và chỉ khi

\(\left(x-\frac{5}{7}\right)^2=0\)

\(=>x-\frac{5}{7}=0\)

\(=>x=\frac{5}{7}\)

vậy GTLN cảu biểu thức là \(-\frac{14087}{7}\) khi và chỉ khi x= \(\frac{5}{7}\)

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+4}{2011}=0\)

\(\Leftrightarrow\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)=0+1+1+1+1\)

\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{2011}=4\)

\(\Leftrightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=4\)

\(\Leftrightarrow x+2015=4:\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)\)

\(\Leftrightarrow x+2015=2012,499379\)

\(\Leftrightarrow x=2012,499379-2015\)

\(\Leftrightarrow x=-2,5006118\)

P/s : Số xấu quá !

<=> \(\frac{\left(x+2014\right)}{2011}+1+\frac{\left(x+2013\right)}{2012}+1=\frac{\left(x+2012\right)}{2013}+1+\frac{\left(x+2011\right)}{2014}+1\)

\(\Rightarrow\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}=\frac{\left(x+4025\right)}{2013}+\frac{\left(x+4025\right)}{2014}\)

=> \(\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}-\frac{\left(x+4025\right)}{2013}-\frac{\left(x+4025\right)}{2014}=0\)

=> \(\left(x+4025\right)\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack=0\)

vì \(\left(\frac{1}{2011}+\frac{1}{2012}\right)>\left(\frac{1}{2013}+\frac{1}{2014}\right)\)

=> \(\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack>0\) hay ≠0

=> \(x+4025=0\)

\(x=-4025\)