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\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}\)
\(x=\frac{23}{11}\)
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
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Xét vế phải
\(\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\frac{9}{1}\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+9\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\left(1+1+...+1\right)\)
\(=\left(1+\frac{1}{9}\right)+\left(1+\frac{2}{8}\right)+\left(1+\frac{3}{7}\right)+\left(1+\frac{4}{6}\right)+\left(1+\frac{5}{5}\right)+\left(1+\frac{6}{4}\right)+\left(1+\frac{7}{3}\right)+\left(1+\frac{8}{2}\right)+1\)\(=\frac{10}{9}+\frac{10}{8}+\frac{10}{7}+\frac{10}{6}+\frac{10}{5}+\frac{10}{4}+\frac{10}{3}+\frac{10}{2}+\frac{10}{10}\)
\(=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
Thay vào bài ,ta được:
\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\times x=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)\(\Rightarrow x=10\)
Vậy x=10
chúc bạn học tốt
a)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+2}=\frac{2}{9}:2\)
\(\frac{1}{x+2}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+2}=\frac{1}{18}\)
=>x+2=18
=>x=16
b tương tự nhân nó với 1/2
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
\(\frac{1}{1\cdot9}+\frac{1}{9\cdot17}+...+\frac{1}{x\left(x+8\right)}=\frac{29}{410}\)
\(\frac{8}{1\cdot9}+\frac{8}{9\cdot17}+...+\frac{8}{x\left(x+8\right)}=\frac{116}{205}\)
\(\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{x}-\frac{1}{x+8}=\frac{116}{205}\)
\(1-\frac{1}{x+8}=\frac{116}{205}\)
\(\frac{x+8}{x+8}-\frac{1}{x+8}=\frac{116}{205}\)
\(\frac{x+7}{x+8}=\frac{116}{205}\)
\(\Leftrightarrow\text{ }205\left(x+7\right)=116\left(x+8\right)\)
\(205x+1435=116x+928\)
\(116x-205x=1435-928\)
\(-89x=507\)
\(x=-5\frac{62}{89}\)
tách vế trái đặt là A\(A=\frac{1}{1.9}+\frac{1}{9.17}+...+\frac{1}{x.\left(x+8\right)}\)
\(A=\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{x}-\frac{1}{x+8}\)
\(A=\frac{1}{1}-\frac{1}{x+8}\)
thay \(\frac{1}{1}-\frac{1}{x+8}=\frac{29}{410}\)
\(\frac{1}{x+8}=\frac{1}{1}-\frac{29}{410}=\frac{410}{410}-\frac{29}{410}\)
\(\frac{1}{x+8}=\frac{381}{410}\)
hình như sai ở đâu