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20 tháng 5 2018
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 \(x\in\left\{0;1;2\right\}\)
25 tháng 7 2018
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
23 tháng 4 2016
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
23 tháng 4 2016
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
4 tháng 8 2018
\(a,\frac{2}{3}.\left(3-x\right)+\frac{1}{2}=\frac{3}{4}.\left(2.x+1\right)
\)
\(2-\frac{2}{3}x+\frac{1}{2}=\frac{3}{2}.\frac{3}{4}x+\frac{3}{4}
\)
\(\frac{2}{3}x+2-\frac{1}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{2}{3}x+\frac{3}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{3}{2}-\frac{3}{4}=\frac{9}{8}x-\frac{2}{3}x\)
\(\frac{6}{4}-\frac{3}{4}=\frac{27}{24}x-\frac{16}{24}x\)
\(\frac{11}{24}x=\frac{3}{4}\)
\(x=\frac{3}{4}:\frac{11}{24}\)
\(x=\frac{3}{4}.\frac{24}{11}\)
\(x=\frac{18}{11}\)
\(Vậy
x=\frac{18}{11}\)
\(b,\frac{5-x}{3}=\frac{2x+1}{5}\)
\(\frac{\left(5-x\right).5}{15}=\frac{\left(2x+1\right).3}{15}\)
\(\Rightarrow\left(5-x\right).5=\left(2x+1\right).3\)
\(25-5x=6x+3\)
\(25-3=6x+5x\)
\(\Rightarrow11x=22\)
\(\Rightarrow x=22:11\)
\(\Rightarrow x=2\)
\(Vậy
x=2\)
3 tháng 8 2018
\(\frac{3}{4}.\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{3}{5}-x=\frac{2}{3}\)
\(x=\frac{3}{5}-\frac{2}{3}\)
\(x=-\frac{1}{15}\)
Vậy \(x=-\frac{1}{15}\)
\(x+\frac{1}{2}.\frac{2}{3}=\frac{3}{4}\)
\(x+\frac{1}{3}=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{1}{3}\)
\(x=\frac{5}{12}\)
vậy \(x=\frac{5}{12}\)
hơ hơ =v
LT
3 tháng 8 2018
\(\frac{3}{4}\times\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{3}{5}-x=\frac{2}{3}\)
\(x=\frac{3}{5}-\frac{2}{3}\)
\(x=\frac{-1}{15}\)
15 tháng 3 2019
1. \(x=\frac{61}{42}\)
2. \(x=\frac{-36}{5}\)
3. \(x=\frac{13}{11}\)
4. \(x=\frac{1}{12}\)
5.\(x=\frac{-5}{2}\)
13 tháng 3 2016
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\frac{100}{100}-\frac{1}{x+1}=\frac{99}{100}\)
\(\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=99\)
DP
13 tháng 7 2017
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
\(\frac{3}{4}.x-1\frac{1}{2}+x=2,4\)
\(\frac{3}{4}.x-\frac{3}{2}+x=2,4\)
\(\frac{3}{4}.x-1.\frac{3}{2}+x.1=2,4\)
\(x.\left(\frac{3}{4}-\frac{3}{2}+1\right)=2,4\)
\(x.\frac{1}{4}=\frac{24}{10}\)
\(x=\frac{24}{10}:\frac{7}{4}\)
\(x.=\frac{24}{10}.\frac{4}{7}\)
\(x=\frac{48}{35}\)
\(\frac{3}{4}x-1\frac{1}{2}+x=2,4\)
<=>\(\frac{3}{4}x-\frac{3}{2}+x-2,4=0\)
,<=>\(\frac{7}{4}x-3,9=0\)
=>\(\frac{7}{4}x=3,9\)
=>\(x=\frac{78}{35}\)