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\(\frac{5}{6}\cdot x+\frac{2}{3}\cdot\left(\frac{3}{5}\cdot x-\frac{6}{5}\right)=\frac{2}{9}-\frac{1}{2}\)
\(\Rightarrow\frac{5}{6}\cdot x+\frac{2}{3}\cdot\frac{3}{5}\cdot x-\frac{2}{3}\cdot\frac{6}{5}=-\frac{5}{18}\)
\(\Rightarrow\frac{5}{6}\cdot x+\frac{2}{5}\cdot x-\frac{4}{5}=-\frac{5}{18}\)
\(\Rightarrow x\cdot\left(\frac{5}{6}+\frac{2}{5}\right)=-\frac{5}{18}+\frac{4}{5}\)
\(\Rightarrow x\cdot\frac{37}{30}=\frac{47}{90}\)
\(\Rightarrow x=\frac{47}{90}\div\frac{37}{30}\)
\(\Rightarrow x=\frac{47}{111}\)
Vậy \(x=\frac{47}{111}\)
1) = \(\frac{3}{5}\)
2) =\(\frac{6}{7}\)
3)\(\frac{9}{13}\)
4)\(\frac{4}{13}\)
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)
\(\Rightarrow\frac{4.4^5}{3.3^5}.\frac{6.6^5}{2.2^5}=2^x\)
\(\Rightarrow\frac{4^6}{3^6}.\frac{6^6}{2^6}=2^x\)
\(\Rightarrow\frac{4^6.6^6}{3^6.2^6}=2^x\)
\(\Rightarrow\frac{\left(4.6\right)^6}{\left(3.2\right)^6}=2^x\)
\(\Rightarrow\frac{24^6}{6^6}=2^x\)
\(\Rightarrow4^6=2^x\)
\(\Rightarrow\left(2^2\right)^6=2^x\)
\(\Rightarrow2^{2.6}=2^x\)
\(\Rightarrow2^{12}=2^x\)
\(\Rightarrow x=12\)
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x.\)
\(\Rightarrow\frac{4.4^5}{3.3^5}.\frac{6.6^5}{2.2^5}=2^x\)\(\Rightarrow\frac{4^6.6^6}{3^6.2^6}=2^x\)
\(\Rightarrow\frac{2^6.2^6.2^6.3^6}{3^6.2^6}=2^x\)\(\Rightarrow2^6.2^6=2^x\)
\(\Rightarrow2^{12}=2^x\Leftrightarrow x=12\)
Bài 1:
a,\(0,75+\frac{9}{17}-1\frac{4}{5}-\frac{26}{17}-2\frac{4}{5}\)
\(=\frac{3}{4}+\left(\frac{9}{17}-\frac{26}{17}\right)-\left(1\frac{4}{5}+2\frac{4}{5}\right)\)
\(=\frac{3}{4}-1-\frac{23}{5}\)
\(=\frac{15}{20}-\frac{20}{20}-\frac{92}{20}=\frac{-97}{20}\)
Bài 2:
a, \(\left(2x+\frac{3}{4}\right)-\frac{10}{3}=\frac{-13}{3}\)
\(2x+\frac{3}{4}=\frac{-13}{3}+\frac{10}{3}\)
\(2x+\frac{3}{4}=-1\)
\(2x=-1-\frac{3}{4}\)
\(2x=\frac{-7}{4}\)
x = -7/8
b, 3,2x - 2,7x + 8,5 = 6
x(3,2 - 2,7) = -2,5
0,5x = -2,5
x = -5
`@`Bảng tần số:
\begin{array}{|c|c|c|}\hline \text{Giá trị (x)}&2&3&4&5&6&7&8&9&10&\\\hline \text{Tần số (n)}&3&5&4&4&4&3&3&2&1&N=29\\\hline\end{array}
`@` Mốt của dấu hiệu là: `3 ( n = 5)`
a) 32.x+2=1342176728
32.x=134217728-2
32.x=134217726
x=134217726:32
x=4194303,938
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