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1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
a) Ta có: \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|2y-1\right|\ge0\end{cases}}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|2y-1\right|+11\ge11\)
\(\Rightarrow A\ge11\)
\(\Rightarrow\)GTNN của A là 11 \(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|2y-1\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}}\)
Vậy ...
b) Ta có: \(\hept{\begin{cases}\left|x-1,2\right|\ge0\\\left|y+1\right|\ge0\end{cases}}\)
\(\Rightarrow\left|x-1,2\right|+\left|y+1\right|+1\ge1\)
\(\Rightarrow\frac{1}{\left|x-1,2\right|+\left|y+1\right|+1}\le1\)
\(\Rightarrow\frac{7}{\left|x-1,2\right|+\left|y+1\right|+1}\le7\)
\(\Rightarrow B\le7\)
\(\Rightarrow\)GTNN của B là 7 \(\Leftrightarrow\hept{\begin{cases}\left|x-1,2\right|=0\\\left|y+1\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1,2\\y=-1\end{cases}}\)
Vậy ...
Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)
\(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)
\(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)
\(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)
Vay \(\frac{A}{B}=\frac{1}{2009}\)
f(x)=9x3-1/3x+3x2-3x+1/3x2-1/9x3-3x2-9x+27+3x
= 9x3-1/9x3+3x2+1/3x2-3x2-1/3-3x-9x+3x+27
= 80/9x3+1/3x2-28/3x+27
\(\frac{1}{1-\frac{2}{1-\frac{3}{1-\frac{1}{4}}}}=\frac{1}{1-\frac{2}{1-\frac{3}{\frac{3}{4}}}}=\frac{1}{1-\frac{2}{1-4}}=\frac{1}{1-\frac{2}{-3}}=\frac{1}{\frac{5}{3}}=\frac{3}{5}\Rightarrow A=1-\frac{3}{5}=\frac{2}{5}\)
Bài làm
\(A=1-\frac{1}{1-\frac{2}{1-\frac{3}{1-\frac{1}{4}}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-\frac{3}{\frac{4}{4}-\frac{1}{4}}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-\frac{3}{\frac{3}{4}}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-3:\frac{3}{4}}}\)
\(A=1-\frac{1}{1-\frac{2}{1-4}}\)
\(A=1-\frac{1}{1-\frac{2}{-3}}\)
\(A=1-\frac{1}{1+\frac{2}{3}}\)
\(A=1-\frac{1}{\frac{3}{3}+\frac{2}{3}}\)
\(A=1-\frac{1}{\frac{5}{3}}\)
\(A=1-1:\frac{5}{3}\)
\(A=1-\frac{3}{5}\)
\(A=\frac{5}{5}-\frac{3}{5}\)
\(A=\frac{2}{5}\)
Vậy \(A=\frac{2}{5}\)
# Học tốt #