\(\left(\frac{1}{25.26}+\frac{1}{26.27}+.......\frac{1}{29.30}\right).150+1,03:\left[1,03.\left(x-1\right)\right]=22\)
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NT
0
9 tháng 8 2016
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)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)
\(\Rightarrow x=4\)
NM
5
KT
11 tháng 7 2018
\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)
\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Leftrightarrow\)\(1+3y=-4\)
\(\Leftrightarrow\)\(3y=-5\)
\(\Leftrightarrow\)\(y=-\frac{5}{3}\)
Vậy...
AK
11 tháng 7 2018
Ta có :
\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y.3=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y.3=-4\)
\(\Rightarrow\left(\frac{5}{120}-\frac{4}{120}\right).120+y.3=-4\)
\(\Rightarrow\frac{1}{120}.120+y.3=-4\)
\(\Rightarrow1+y.3=-4\)
\(\Rightarrow3y=-4-1\)
\(\Rightarrow3y=-5\)
\(\Rightarrow y=-\frac{5}{3}\)
Vậy \(y=-\frac{5}{3}\)
