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\(x+\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{9}{2}\)
\(x+\frac{3\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{9}{2}\)
\(x+\frac{3}{5}=\frac{9}{2}\)
\(x=\frac{39}{10}\)
\(x=3,9\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
\(\left(a\right)\frac{34-x}{30}=\frac{5}{6}\)
\(\frac{34-x}{30}=\frac{25}{30}\)
34 - x = 25
x = 34 - 25 = 9
\(\left(b\right)\frac{x+13}{34}=\frac{12}{17}\)
\(\frac{x+13}{34}=\frac{24}{34}\)
x + 13 = 24
x = 24 - 13 = 11
\(\left(c\right)\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{56}{81}\)
\(4x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Ta có : \(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)
\(2A=1-\frac{1}{81}=\frac{80}{81}\)
\(A=\frac{80}{81}\div2=\frac{40}{81}\)
\(\Rightarrow4x+\frac{40}{81}=\frac{56}{81}\)
\(4x=\frac{56}{81}-\frac{40}{81}\)
\(4x=\frac{16}{81}\)
\(x=\frac{16}{81}\div4=\frac{4}{81}\)
a) 15 + 3 ( x - 1 ) = 36
3 ( x - 1 ) = 36 - 15
3 ( x - 1 ) = 21
x - 1 = 21 : 3
x - 1 = 7
x = 7+1
x = 8
\(\frac{2}{3}x+\frac{10}{9}=\frac{23}{18}\)
\(\frac{2}{3}x=\frac{23}{18}-\frac{10}{9}\)
\(\frac{2}{3}x=\frac{1}{6}\)
\(x=\frac{1}{6}:\frac{2}{3}\)
\(x=\frac{1}{4}\)
\(\frac{x+9}{13-x}=\frac{6}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+9=6\\13-x=5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
\(\frac{120-x}{30}=3\frac{1}{2}\)
\(\Rightarrow\frac{120-x}{30}=\frac{7}{2}\)
\(\Rightarrow\frac{120-x}{30}=\frac{105}{30}\)
\(\Rightarrow120-x=105\)
\(\Rightarrow x=120-105\)
\(\Rightarrow x=15\)