(9,1+6,9) x 1\(\frac{4}{4}\): 2
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\(=4,85-\left(\frac{25}{8}+1,105\right)4,85-\left(3,125+1,105\right)=4,85-4,23=0,62\)
4,58 - (3,125 + 1,105) < x < 9,1 - ( 6,85 - 2,75)
=> 4,58 - 4,23 < x < 9,1 - 4,1
=> 0,35 < x < 5
=> x \(\in\){1; 2; 3; 4}
4.58-(3.125+1.105)<x<9.1-(6.85-2.75)
4.58-4.28<x<9.1-4.1
0.35<x<5
x\(\in\)1.2.3.4
\(x:\frac{13}{10}+\frac{42}{5}.\frac{6}{7}.\left(6-\frac{\left(2,3+0,8\right).7}{0,1+6,9}\right)=\frac{39}{7}:\frac{15}{14}=\frac{37}{7}.\frac{14}{15}=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{36}{7}\left(6-\frac{3,1.7}{7}\right)=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{36}{7}\left(6-3,1\right)=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{36}{7}.\frac{29}{10}=\frac{74}{15}\)
\(x.\frac{10}{13}+\frac{522}{35}=\frac{74}{15}\)
\(x.\frac{10}{13}=\frac{74}{15}-\frac{522}{35}=-9\frac{103}{105}\)
\(x=-9\frac{103}{105}:\frac{10}{13}=-12\frac{512}{525}\)
\(5\frac{4}{7}\): [ x : 1,3 + 8,4 . \(\frac{6}{7}\). ( 6 - \(\frac{\left(2,3+5\div6,25\right)\times7}{8\times0,-125+6,9}\)) ] = \(1\frac{1}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{\left(2,3+0,8\right).7}{0,1+6,9}\)) ] = \(\frac{15}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{3,1.7}{7}\)) ] = \(\frac{15}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - 3,1 ) ] = \(\frac{15}{14}\)
x : 1,3 + \(\frac{36}{5}\). 2,9 = \(\frac{39}{7}\): \(\frac{15}{14}\)
x : 1,3 + 20,88 = 5,2
x : 1,3 = - 15,68
x = - 15,68 . 1,3
x = - 20,384
ta có
\(5\frac{4}{7}:\left\{x:1,3+8,4.\frac{6}{7}.\left[6-\frac{\left(2,3+5:6,25\right).7}{8.0,0125+6,9}\right]\right\}=1\frac{1}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{\left(2,3+0,8\right).7}{0,1+6,9}\right]\right\}=\frac{15}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{3,1.7}{7}\right]\right\}=\frac{15}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.2,9\right\}=\frac{15}{14}\Leftrightarrow\left\{x:1,3+7,2.2,9\right\}=\frac{39}{7}:\frac{15}{14}\)
\(\Leftrightarrow x:1,3+20,88=5,2\Leftrightarrow x:1,3=-15,68\Leftrightarrow x=-20,384\)
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