\(\text{Tìm GTNN, GTLN của biểu thức }:\)
\(P=\sqrt{x-5}+\sqrt{7-x}\)
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\(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\)
\(=\frac{1}{2}.\frac{2}{97.99}-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\)
\(=\frac{1}{2}.\left[\frac{2}{97.99}-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\right]\)
\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\right]\)
\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{97}\right)\right]\)
\(=\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}-\frac{98}{97}\right)\)
\(=\frac{1}{2}.\left(-1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{-100}{99}\)
\(=-\frac{50}{99}\)
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) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)
\(\Rightarrow x-1,01=1\)
\(\Rightarrow x=1+1,01\)
\(\Rightarrow x=2,01\)
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!
ĐKXĐ:
\(\sqrt{x-5}\ge0\Rightarrow x\ge5\)
\(\sqrt{7-x}\ge0\Rightarrow x\le7\)
=> Pmax =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..........