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a) \(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(\Rightarrow x+\frac{7}{12}=\frac{5}{6}\)
\(\Rightarrow x=\frac{5}{6}-\frac{7}{12}\)
\(\Rightarrow x=\frac{1}{4}\)
b) \(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\Rightarrow\frac{13}{23}+x=\frac{29}{30}-\frac{7}{69}\)
\(\Rightarrow\frac{13}{23}+x=\frac{199}{230}\)
\(\Rightarrow x=\frac{199}{230}-\frac{13}{23}\)
\(\Rightarrow x=\frac{3}{10}\)
a)\(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(x+\frac{7}{12}=\frac{5}{6}\)
\(x=\frac{5}{6}-\frac{7}{12}\)
\(x=\frac{1}{4}\)
b)\(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\left(\frac{13}{23}+x\right)=\frac{29}{30}-\frac{7}{69}\)
\(\left(\frac{13}{23}+x\right)=\frac{199}{230}\)\(x=\frac{199}{230}-\frac{13}{23}\)
\(x=\frac{3}{10}\)
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
Bài 1: Tìm x biết:
1) x +\(\frac{7}{12}\)= \(\frac{17}{18}\)- \(\frac{1}{9}\) 2) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+ x) = \(\frac{7}{69}\)
x +\(\frac{7}{12}\)= \(\frac{15}{18}\) \(\frac{13}{23}\)+ x = \(\frac{29}{30}\)- \(\frac{7}{69}\)
x = \(\frac{15}{18}\)- \(\frac{7}{12}\) \(\frac{13}{23}\)+ x = \(\frac{199}{230}\)
x = \(\frac{1}{4}\) x = \(\frac{3}{10}\)
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow2^x\cdot1+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=480\)
\(\Rightarrow2^x\left(1+2^1+2^2+2^3\right)=480\)
\(\Rightarrow2^x\cdot15=480\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(\frac{2011}{2}+1\right)+...+\left(\frac{2}{2011}+1\right)+\left(\frac{1}{2012}+1\right)+1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow x=2013.\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}\)
\(\Rightarrow x=2013\)
Vậy \(x=2013\)
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
a) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+X)=\(\frac{7}{69}\)
\(\frac{13}{23}\)+X=\(\frac{29}{30}\)-\(\frac{7}{69}\)
\(\frac{13}{23}\)+X=\(\frac{199}{230}\)
X=\(\frac{199}{230}\)-\(\frac{13}{23}\)
X=\(\frac{3}{10}\)
b)1/2+1/6+1/12+...+1/x(x+1)=2011/2012
=>1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2011/2012
=>1-1/2+1/2-1/3+1/3+1/4+...+1/x+1/x+1=2011/2012
=>1-1/x+1=2011/2012
=>1/x+1=1-2011-2012
=>1/x+1=2012/2012-2011/2012
1/x+1=1/2012
=>x+1=2012
=>x=2011
a) 3/10