ĐKXĐ:

 \(\sqrt{x-5}\ge0\Rightarrow x\ge5\)

\(\sqrt{7-x}\ge0\Rightarrow x\le7\)

=> P_max =2 tại x=7

DKXD:\(5\le x\le7\)

GTLN: \(P=\sqrt{x-5}+\sqrt{7-x}=1.\sqrt{x-5}+1.\sqrt{7-x}\)

                                  \(\le\frac{1^2+\left(\sqrt{x-5}\right)^2}{2}+\frac{1^2+\left(\sqrt{7-x}\right)^2}{2}\left(bdtCOSI\right)\)

                                    \(=\frac{2+x-5+7-x}{2}=2\)

                       "="\(\Leftrightarrow\hept{\begin{cases}1=\sqrt{x-5}\\1=\sqrt{7-x}\\7\ge x\ge5\end{cases}}\Leftrightarrow x=6\)

Vậy..............................................................

GTNN: ta sẽ chứng minh: \(P\ge\sqrt{2}\)

 bđt có thể viết lại thành:\(\sqrt{x-5}+\sqrt{7-x}\ge\sqrt{2}\Leftrightarrow\left(\sqrt{x-5}+\sqrt{7-x}\right)^2\ge\left(\sqrt{2}\right)^2\)

                                       \(\Leftrightarrow x-5+7-x+2\sqrt{\left(x-5\right)\left(7-x\right)}\ge2\Leftrightarrow2+2\sqrt{\left(x-5\right)\left(7-x\right)}\ge2\)

                                       \(\Leftrightarrow2\sqrt{\left(x-5\right)\left(7-x\right)}\ge0\)(đúng với mọi x thỏa mãn \(7\ge x\ge5\))

          "="\(\Leftrightarrow\hept{\begin{cases}2\sqrt{\left(x-5\right)\left(7-x\right)}\\7\ge x\ge5\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=7\end{cases}}}\)

                      Vậy..........

\(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{3.1}\)

\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\)

\(=\frac{1}{2}.\frac{2}{97.99}-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\)

\(=\frac{1}{2}.\left[\frac{2}{97.99}-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\right]\)

\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\right]\)

\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{97}\right)\right]\)

\(=\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}-\frac{98}{97}\right)\)

\(=\frac{1}{2}.\left(-1-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\frac{-100}{99}\)

\(=-\frac{50}{99}\)

1) Tính : 

a) \(\left(2008.2009.2010.2011\right).\left(1+\frac{1}{2}:\frac{2}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)

\(=\left(2008.2009.2010.2011\right).0\)

\(=0\)

2) Tìm x 

a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2012\)

b) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)

\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)

\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)

\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)

\(\Rightarrow x-1,01=1\)

\(\Rightarrow x=1+1,01\)

\(\Rightarrow x=2,01\)

1400958

1 400 958

mình hog bít

Thay \(z=x+y+1\) vào P ta có:

\(P=\frac{x^3y^3}{\left\{\left[x+y\left(x+y+1\right)\right]\left[y+x\left(x+y+1\right)\right]\left[xy+y+x+z\right]\right\}^2}\)

    \(=\frac{x^3y^3}{\left[\left(x+1\right)\left(y+1\right)\left(x+y\right)^2\right]^2}\)

Mà \(x+1\ge2\sqrt{x};y+1\ge2\sqrt{y};x+y\ge2\sqrt{xy}\)

=> \(P\le\frac{x^3y^3}{\left(2\sqrt{x}.2\sqrt{y}.4xy\right)^2}=\frac{1}{256}\)

MaxP=1/256  khi \(a=b=1;c=3\)

\(-\frac{7}{12}+\frac{11}{8}-\frac{5}{9}\)

= \(-\frac{42}{72}+\frac{99}{72}-\frac{40}{72}\)

= \(\frac{17}{72}\)

Chúc bạn học tốt !!!

\(-\frac{7}{12}+\frac{11}{8}-\frac{5}{9}\)

\(=\frac{-14+33}{24}-\frac{5}{9}\)

\(=\frac{19}{24}-\frac{5}{9}=\frac{57-40}{72}=\frac{17}{72}\)

                                   Số các số hạng là : 

                                     \(\left(25,25-1,25\right):1,5+1=17\)

                                Tổng dãy số trên là :

                                \(\left(25,25+1,25\right).17:2=225,25\)

                   Chúc bạn học tốt !!!

Ta có:1,25 - 2,75 + 4,25 - 5.75 + ... + 25,25

      =>\(\frac{5}{4}-\frac{11}{4}+\frac{17}{4}-\frac{23}{4}+...+\frac{101}{4}\)

      =>\(\frac{5}{4}+\frac{17}{4}+....+\frac{101}{4}-\frac{11}{4}-\frac{23}{4}-....-\frac{95}{4}\)

      =>\(\left(\frac{5}{4}+\frac{17}{4}+...+\frac{101}{4}\right)-\left(\frac{11}{4}+\frac{23}{4}+....+\frac{95}{4}\right)\)

      =>\(\left(\frac{5+17+...+101}{4}\right)-\left(\frac{11+23+....+95}{4}\right)\)

      =>\(\left(\frac{\left(101+5\right).\left[\left(101-5\right):12+1\right]:2}{4}\right)-\left(\frac{\left(95+11\right).\left[\left(95-11\right):12+1\right]:2}{4}\right)\)

      =>\(\left(\frac{106.\left(96:12+1\right):2}{4}\right)-\left(\frac{106.\left(84:12+1\right):2}{4}\right)\)

      =>\(\left(\frac{53.\left(8+1\right)}{4}\right)-\left(\frac{53.\left(7+1\right)}{4}\right)\)

     =>\(\left(\frac{53.9}{4}\right)-\left(\frac{53.8}{4}\right)\)

     =>\(\frac{477}{4}-\frac{424}{4}=\frac{53}{4}=13,25\)

Vậy 1,25 - 2,75 + 4,25 - 5.75 + ... + 25,25 = 13,25

Chúc bn học tốt!

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