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= \(\frac{5}{7}\)

\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)

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

\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)

\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)

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

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

Làm nốt

\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)

\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2013}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

tự làm nốt~

kudo shinichi làm sai ở chỗ:

\(\frac{x-2013}{2011}\)phải là \(\frac{x-2014}{2011}\)mới đúng nhé

Ngọc Nguyễn: uk. mình gõ nhầm

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

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)

\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0

\(\Leftrightarrow x=2013\)

Vậy ........

Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)

\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)

\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)

\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)

=> 2015 + x = 0

=> x = -2015

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k biet nen k tra loi

tham khảo Câu hỏi của Đỗ Thu Hà - Toán lớp 9 - Học toán với OnlineMath

ĐKXĐ:x≠2009;y≠2010;z≠2011;x,y,z∈R

√x−2009−1x−2009 +√y−2010−1y−2010 +√z−2011−1z−2011 =3/4⇔1x−2009 −√x−2009x−2009 +1y−2010 −√y−2011y−2011 +1z−2011 −√z−2011z−2011 =−34⇔(1√x−20092 −1√x−2009 +14 )+(1√y−20102 −1√y−2010 +14 )+(1√z−20112 +1√z−2011 +14 )=0⇔(1√x−2009 −12 )2+(1√y−2010 −12 )2+(1√z−2011 −12 )2=0

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