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a)\(\dfrac{3.7.13.37.39-10101}{505050-70707}\)=\(\dfrac{10101.39-10101}{10101.50-10101.7}\)=\(\dfrac{10101.\left(39-1\right)}{10101.\left(50-7\right)}\)=\(\dfrac{39-1}{50-7}\)=\(\dfrac{38}{43}\)
b)\(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\)=\(\dfrac{36.17+36.\left(-62\right)}{36.\left(-17\right)+36.\left(-13\right)}\)=
\(\dfrac{36.\left[17+\left(-62\right)\right]}{36.\left[\left(-17\right)+\left(-13\right)\right]}\)=\(\dfrac{36.\left(-45\right)}{36.\left(-20\right)}\)=\(\dfrac{45}{20}\)=\(\dfrac{9}{4}\)=2\(\dfrac{1}{4}\)
a) \(\frac{2929-101}{2.1919+404}\)\(=\frac{101.29-101}{2.1919+2.202}\)\(=\frac{101\left(29-1\right)}{2\left(1919+202\right)}\)\(=\frac{101.28}{2.2121}\)\(=\frac{2.101.7.2}{2.101.7.3}\)\(=\frac{2}{3}\)
b) \(\frac{-1997.1996+1}{-1995\left(-1997\right)+1996}\)\(=\frac{-1997.\left(1995+1\right)+1}{-1995\left(-1997\right)+1996}\)\(=\frac{-1997.1995-1997+1}{-1\left(-1997\right).1995+1996}\)\(=\frac{-1997.1995-1996}{1997.1995+1996}\)\(=\frac{-\left(1997.1995+1996\right)}{1997.1995+1996}\)\(=-1\)
c) \(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}\)\(=\frac{2.3+4.2.3+2.7.3.7}{3.5+2.3.2.5+3.7.5.7}\)\(=\frac{2.3\left(1+4+7.7\right)}{3.5\left(1+2.2+7.7\right)}\)\(=\frac{2\left(1+4+49\right)}{5\left(1+4+49\right)}\)\(=\frac{2}{5}\)
d) \(\frac{3.7.13.37.39-10101}{505050-70707}\)\(=\frac{3.7.13.37.39-3.7.13.37}{2.3.5.5.7.13.37-3.7.7.13.37}\)\(=\frac{3.7.13.37\left(39-1\right)}{3.7.13.37\left(2.5.5-7\right)}\)\(=\frac{38}{43}\)
e) \(\frac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\)\(=\frac{18.2.17-18.2.62}{\left(-1\right).2.2.3.3.17+\left(-1\right)2.2.3.3.13}\)\(=\frac{18.2\left(17-62\right)}{\left(-1\right).2.2.3.3\left(17+13\right)}\)\(=\frac{\left(-1\right).2.2.3.3.45}{\left(-1\right).2.2.3.3.30}\)\(=\frac{45}{30}\)\(=\frac{3}{2}\)
18,34+(-18)x124=18,34+-2232
=2213,66
-36,17+9 x-52= -36,17+468
=431,83
a) \(\dfrac{18.34-18.124}{-36.17+9.\left(-4\right).13}=\dfrac{18.\left(34-124\right)}{-36.\left(17+13\right)}=\dfrac{18.\left(-90\right)}{36.\left(-30\right)}=\dfrac{1.\left(-3\right)}{2.\cdot\left(-1\right)}=\dfrac{3}{2}\)
b) \(\dfrac{393939-10101}{505050-70707}=\dfrac{10101.39-10101}{10101.50-10101.7}=\dfrac{10101.\left(39-1\right)}{10101.\left(50-7\right)}=\dfrac{38}{43}\)
a) \(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}=\dfrac{36.17+36.\left(-62\right)}{36.\left(-17\right)+35.\left(-13\right)}=\dfrac{36\left[17+\left(-62\right)\right]}{36\left[\left(-17\right)+\left(-13\right)\right]}=\dfrac{45}{20}=\dfrac{9}{4}\)
b) \(\dfrac{3.7.13.37.39-10101}{505050-70707}=\dfrac{10101.39-10101}{10101.50-10101.7}=\dfrac{10101\left(39-1\right)}{10101\left(50-7\right)}=\dfrac{38}{43}\)
\(\frac{13.\left(3-8\right)}{40.\left(15-2\right)}=\frac{13.\left(-5\right)}{40.13}=\frac{-1}{8}\)
\(\frac{18.34+\left(-18\right).124}{\left(-36\right).17+9.\left(-52\right)}=\frac{18.34+18.\left(-124\right)}{\left(-36\right).17+9.\left(-4\right).13}=\frac{18.\left[34+\left(-124\right)\right]}{\left(-36\right).\left(17+13\right)}=\frac{18.\left(-90\right)}{\left(-36\right).30}=\frac{3}{2}\)
cho minh dung di
\(C=\frac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}=\frac{18.\left(34-124\right)}{9.\left(-4.17-52\right)}=\frac{18.\left(-90\right)}{9.\left(-120\right)}=\frac{3}{2}\)
\(c=\frac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\)
\(=\frac{18.34+18.\left(-124\right)}{9.-3.17+9.\left(-52\right)}\)
\(=\frac{18.\left(34+\left(-124\right)\right)}{9.\left(-3+17\right).\left(52\right)}\)
tk ủng hộ mk nha!!!!!!!!!!