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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)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
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)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)\)
A = \(\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{2004}{2004}-\frac{1}{2004}\right)\)
A = \(\frac{1}{2}\)x\(\frac{2}{3}.\)\(\frac{3}{4}....\)\(\frac{2003}{2004}\)
A = \(\frac{1}{2004}\)
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
Bài 1:
a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)
\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)
\(=\frac{2+3}{12}=\frac{5}{12}\)
b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{-7}{10}\)
\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)
Bài 2:
a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)
\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)
\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)
\(\Leftrightarrow-2x=-6-1=-7\)
hay \(x=\frac{7}{2}\)
Vậy: \(x=\frac{7}{2}\)
4a) \(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)
\(\Leftrightarrow x=\frac{1}{10}:\frac{-2}{3}=\frac{1}{10}.\frac{3}{-2}=\frac{3}{-20}\)
Vậy x=\(\frac{3}{-20}\)
b) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\Leftrightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)
\(\Leftrightarrow\frac{-5}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{12}:\frac{-5}{6}=\frac{5}{12}.\frac{6}{-5}=\frac{1}{-2}\)
Vậy x=\(\frac{1}{-2}\)
g)Sửa đề: \(\left|4x-1\right|=\left(-3\right)^2\)
\(\Leftrightarrow\left|4x-1\right|=9\)
\(\Rightarrow\left[{}\begin{matrix}4x-1=9\\4x-1=\left(-9\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{2};-2\right\}\)
i) \(\left(x-1^3\right)=125\)
\(\Leftrightarrow x-1=125\)
\(\Leftrightarrow x=125+1=126\)
Vậy x=126
k) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
tích mình đi
ai tích mình
mình tích lại
thanks
k mk đi