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ta có:
\(1+\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{x\left(x+3\right)}\)=1\(\frac{7}{24}\)\(=\frac{31}{24}\)
\(1+\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}=\frac{31}{24}\)
\(3+\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{31}{3}\)
\(3+1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{31}{3}\)
\(3+1-\frac{1}{x+3}=\frac{31}{3}\)
\(4-\frac{1}{x+3}=\frac{31}{3}\)
\(\frac{1}{x+3}=4-\frac{31}{3}=\frac{-19}{3}\)
=>ko tìm được x
A, \(2\frac{2}{5}\left(\frac{1}{2}x-0,75\right)=\frac{3}{10}\)
\(=>\frac{2.5+2}{5}\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(=>\frac{1}{2}x-\frac{3}{4}=\frac{3}{10}:\frac{12}{5}=\frac{1}{8}\)
\(=>x=\left(\frac{1}{8}+\frac{3}{4}\right):\frac{1}{2}\)
\(=>x=\frac{7}{4}\)
B, \(\frac{3}{5}-|x-\frac{1}{2}|=25\%\)
\(=>|x-\frac{1}{2}|=\frac{3}{5}-\frac{1}{4}\)
\(=>|x-\frac{1}{2}|=\frac{7}{20}\)
\(=>x-\frac{1}{2}=\frac{7}{20};-\frac{7}{20}\)
TH1: \(x-\frac{1}{2}=\frac{7}{20}=>x=\frac{17}{20}\)
TH2: \(x-\frac{1}{2}=-\frac{7}{20}=>x=\frac{3}{20}\)
Bài giải:
Câu 1: a, \(\left(-2\right).4.5.38.\left(-25\right)\)
\(=\left[\left(-2\right).5\right].\left[4.\left(-25\right)\right].38\)
\(=\left(-10\right).\left(-100\right).38\)
\(=1000.38=38000\)
b,\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\)
\(=\left(\frac{1}{3}+\frac{3}{8}\right)-\frac{7}{12}\)
\(=\frac{17}{24}-\frac{7}{12}=\frac{1}{8}\)
c, \(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}\)
\(=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}\)
\(=\frac{-5}{8}.1+\frac{17}{8}\)
\(=\frac{3}{2}\)
Câu 2: a, \(x-\frac{2}{5}=0,24\)
\(x-0,4=0,24\)
\(x=0,24+0,4\)
\(\Rightarrow x=0,64\left(\frac{16}{25}\right)\)
b,\(\frac{2}{3}.x+\frac{1}{12}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{12}\)
\(\frac{2}{3}.x=\frac{1}{60}\)
\(x=\frac{1}{60}:\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{40}\)
c, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)
\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)
\(\frac{7}{2}-2x=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=-2\)
\(\Rightarrow x=-2:2\)
\(x=-1\)
Câu I: Ta có:
|5 - 3x| + 2/3=1/6
\(\Rightarrow\) |5 - 3x| =1/6- 2/3
\(\Rightarrow\) |5 - 3x| =-1/2
\(\Rightarrow\)5-3x= -1/2 hoặc 5-3x=1/2
\(\Rightarrow\)x=11/6 hoặc x=3/2.
Câu K: Ta có
- 2,5 + |3x + 5| = -1,5
\(\Rightarrow\) |3x + 5| = -1,5-(- 2,5 )
\(\Rightarrow\) |3x + 5| = 1
\(\Rightarrow\)3x + 5 = 1 hoặc 3x + 5 = -1
\(\Rightarrow\)x= -4/3 hoặc x= -2
\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=3+\frac{6}{5}\)
\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)
\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}\)
\(\Rightarrow x=\frac{42}{11}\)
\(b,\left(3\frac{1}{2}-x\right).1\frac{1}{4}=-1\frac{1}{20}\)
\(\Rightarrow\left(\frac{7}{2}-x\right)\cdot\frac{5}{4}=-\frac{21}{20}\)
\(\Rightarrow\frac{7}{2}-x=-\frac{21}{20}:\frac{5}{4}\)
\(\Rightarrow\frac{7}{2}-x=-\frac{21}{25}\)
\(\Rightarrow x=\frac{7}{2}--\frac{21}{25}\)
\(\Rightarrow x=\frac{7}{2}+\frac{21}{25}\)
\(\Rightarrow x=\frac{217}{50}\)