Tìm x biết : \(1\frac{2}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right)\times x=\frac{16}{5}\)
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\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right)x=\frac{16}{5}\)
\(\Rightarrow\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}.x=\frac{16}{5}-\frac{8}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=4\)
vậy x=4
=.8/5+((2*17*37+2*7*37+2*7*17)/(7*17*37))/((5*17*37+5*7*37+5*7*17)/(7*17*37))*x=16/5
=>8/5+((2*17*37+2*7*37+2*7*17)/(7*17*37))*((7*17*37)/(5*17*37+5*7*37+5*7*17))*x=16/5
=>8/5+(2(17*37+7*37+7*17))/(5(17*37+7*37+7*17))*x=16/5
=>8/5+(2/5)*x=16/5
=>(2/5)*x=16/5-8/5
=>(2/5)*x=8/5
=>x=(8/5)/(2/5)
=>x=4
Vậy x=4
a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(\Rightarrow1+x:\frac{1}{3}=-4\)
\(\Rightarrow x:\frac{1}{3}=-4-1=-5\)
\(\Rightarrow x=-5.\frac{1}{3}=\frac{-5}{3}\)
b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)
\(\Rightarrow x=4\)
1.
A=\(\frac{-5x+-5y+-5z}{21}=\frac{-5\left(x+y+z\right)}{21}=\frac{-5}{21}.x+y+z\)
A= -z+z=0
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a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
\(\frac{7}{5}+\left(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right)\cdot x=\frac{16}{5}\)
\(\frac{2}{5}x=\frac{16}{5}-\frac{7}{5}\)
\(\frac{2}{5}x=\frac{9}{5}\)
x = \(\frac{9}{5}:\frac{2}{5}\)
x = 9/2
\(1\frac{2}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)
\(\frac{7}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)
\(\frac{7}{5}+\frac{2}{5}.x=\frac{16}{5}\)
\(\frac{2}{5}.x=\frac{16}{5}-\frac{7}{5}\)
\(\frac{2}{5}.x=\frac{9}{5}\)
\(x=\frac{9}{5}:\frac{2}{5}\)
\(x=\frac{9}{5}.\frac{5}{2}\)
\(x=\frac{9}{2}\)