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\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{-1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}\)
\(=\frac{16}{35}\)
\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{7.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(A=\frac{3}{5}+\frac{1}{7}=\frac{21}{35}+\frac{5}{35}=\frac{26}{35}\)
B=\(\frac{12+\frac{4}{3}-\frac{12}{37}-\frac{12}{53}}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12+\frac{12}{9}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{9}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=\(\frac{12\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{1}+\frac{1}{9}-\frac{1}{37}-\frac{1}{53}\right)}:\frac{4\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(\frac{1}{1}+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
=\(4:\frac{4}{5}\)
=\(5\)
\(A=\frac{636363.37-373737.63}{1+2+3+...+2006}\)
\(A=\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
\(A=0\)
\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left[\frac{4.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}{1.\left(3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}\right)}:\frac{4.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}{5.\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}\right)}\right].\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\left(4:\frac{4}{5}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.5.\frac{124242423}{237373735}\)
\(B=\frac{47.5.124242423}{41.237373735}\)
\(B=\frac{29196969405}{9732323135}\)
Ủng hộ mk nha !!! ^_^
a) \(A=\frac{636363.37-373737.63}{1+2+3+...+2006}\)
\(A=\frac{10101.63.37-10101.37.63}{1+2+3+...+2006}\)
\(A=\frac{0}{1+2+3+...+2006}\)
\(A=0\)
b) \(B=1\frac{6}{41}\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}.\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
\(B=\frac{47}{41}.\frac{12}{3}.\left(\frac{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}{1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}}\right).\frac{4}{5}.\left(\frac{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}{1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2006}}\right).\frac{123}{235}\)
\(B=\frac{47.4.4.123}{41.5.235}\)
\(B=\frac{47.4.4.41.3}{41.5.47.5}\)
\(B=\frac{4.4.3}{5.5}\)
\(B=\frac{48}{25}\)
\(P=-1\frac{1}{5}.\frac{4\left(3\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
=> \(P=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
=> \(P=-1\frac{1}{5}.4:\frac{4}{5}\)
=> \(P=-\frac{6}{5}.4.\frac{5}{4}=-6\)
P/s : Đề của bạn sai nên mik đã sửa lại rồi
Ta có :
\(B=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(\Rightarrow B=-\frac{6}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{1\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(\Rightarrow B=-\frac{6}{5}.4:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}.\frac{5}{4}\)
\(\Rightarrow B=-6\)
Vậy \(B=-6\)
~ Ủng hộ nhé
Bài này về tỷ số
Khó lắm bạn ơi
Thông cảm cho mình
~~~ Chúc bạn học giỏi ~~~
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)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)
\(\Rightarrow x=4\)
chịu
<=> A=\(\frac{3\left(\frac17-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac17-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\left(\frac12-\frac13+\frac14-\frac15\right)}{-7\left(\frac12-\frac13+\frac14-\frac15_{}\right)}\)
\(A=\frac35+\left(-\frac17\right)\)
\(A=\frac{16}{35}\)
Ta có $A = \dfrac{\dfrac{3}{7} - \dfrac{3}{17} + \dfrac{3}{37}}{\dfrac{5}{7} - \dfrac{5}{17} + \dfrac{5}{37}} + \dfrac{\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5}}{\dfrac{7}{5} - \dfrac{7}{4} + \dfrac{7}{3} - \dfrac{7}{2}}$
$= \dfrac{3 \cdot \left(\dfrac{1}{7} - \dfrac{1}{17} + \dfrac{1}{37}\right)}{5 \cdot \left(\dfrac{1}{7} - \dfrac{1}{17} + \dfrac{1}{37}\right)} + \dfrac{\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5}}{- \left(\dfrac{7}{2} - \dfrac{7}{3} + \dfrac{7}{4} - \dfrac{7}{5}\right)}$
$= \dfrac{3}{5} + \dfrac{\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5}}{-7 \cdot \left(\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5}\right)}$
$= \dfrac{3}{5} + \dfrac{1}{-7}$
$= \dfrac{3}{5} - \dfrac{1}{7}$
$= \dfrac{21}{35} - \dfrac{5}{35}$
$= \dfrac{16}{35}$