Tìm x :
\(2\cdot\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
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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 làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)
b)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
Vậy \(x=-329\)
\(\frac{1}{|x-y|\sqrt{x^6\left(x-y\right)^2}}=\frac{1}{|x-y|x^3|x-y|}\)
=\(\frac{1}{x^3\left(x-y\right)^2}\)
a) 4/3 - x = 3/5 + 1/2
=> 4/3 - x= 0,8
=> x = 4/3 + 0/8
=> x = 5/8
ta có:
\(\left(7\frac{1}{2}.8\frac{3}{70}+8\frac{3}{70}.\frac{9}{4}+\frac{19}{4}.8\frac{3}{70}+5\frac{1}{2}.8\frac{3}{70}\right):x=1126\)
\(8\frac{3}{70}\left(7\frac{1}{2}+\frac{9}{4}+\frac{19}{4}+5\frac{1}{2}\right):x=1126\)
\(\frac{563}{70}.\left(\frac{15}{2}+7+\frac{11}{2}\right):x=1126\)
\(\frac{563}{70}.20:x=1126\)
\(\frac{1126}{70}:x=1126\)
\(=>x=\frac{1126}{7}:1126\)
\(=>x=\frac{1}{7}\)
cho mình nha các bạn.
\(2\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(1-2x+\frac{1}{3}=\frac{7}{6}-x\)
\(-2x+x=\frac{7}{6}-\frac{1}{3}-1\)
\(-x=-\frac{1}{6}\)
\(x=\frac{1}{6}\)
Vậy \(x=\frac{1}{6}\)