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\(A=2016:4+4048:2\\ =504+2024\\ =2528\\ B=37064-64\cdot\left(82+42966:2017\right)\\ =37064-64\cdot\left(82+21,30193356\right)\\ =37064-64\cdot103,3019336\\ =37064-6611,323748\\ =30452,67625\\ C=\left(100+92\cdot42\right)+\left(200-58\right)\cdot58\\ =\left(100+3864\right)+142\cdot58\\ =3964+8236\\ =12200\\ D=\left(1998:18-1443:13\right)\cdot\left(16996-1110:30\cdot305\right)\\ =\left(111-111\right)\cdot\left(16996-1110:30\cdot305\right)\\ =0\cdot\left(16996-1110:30\cdot305\right)\\ =0\)
bài 1b)
\(8\frac{1}{14}-6\frac37\)
C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)
C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)
\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)
bài 1 c)\(7-3\frac67\)
C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)
C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)
a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)
\(=\frac{8+9+10}{12}\)
\(=\frac{27}{12}=\frac{9}{4}\)
b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)
\(=\frac{45-14+20}{24}\)
\(=\frac{51}{24}=\frac{17}{8}\)
2)
a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)
\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)
\(=1+\frac{7}{13}+\frac{1}{7}\)
\(=\frac{20}{13}+\frac{1}{7}\)
\(=\frac{153}{91}\)
Tí tớ trả lời tiếp
b)\(5\frac14+3\frac25-4\frac14\)
=\(\left(5\frac14-4\frac14\right)+3\frac25\)
=\(\left\lbrack\left(5-4\right)+\left(\frac14-\frac14\right)\right\rbrack+\frac{17}{5}\)
=\(1+0+\frac{17}{5}\)
=\(\frac55+\frac{17}{5}\)
=\(\frac{22}{5}\)
Bài 2:
- \(\frac75\) + \(x\) = - \(\frac34\)
\(x\) = - \(\frac34+\frac75\)
\(x=-\frac{15}{20}\) + \(\frac{28}{20}\)
\(x\) = \(\frac{13}{20}\)
Vậy \(x=\frac{13}{20}\)
Bài 3:
- \(\frac74\) + \(\frac14:x\) = \(\frac32\)
\(\frac14:x\) = \(\frac32+\frac74\)
\(\frac14:x\) = \(\frac64+\frac74\)
\(\frac14:x=\frac{13}{4}\)
\(x=\frac14:\frac{13}{4}\)
\(x=\frac14\times\frac{4}{13}\)
\(x=\frac{1}{13}\)
Vậy \(x=\frac{1}{13}\)
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
a)\(\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}\)
\(=\frac{-7}{25}.1-\frac{18}{25}\)
\(=-\frac{7}{25}-\frac{18}{25}\)
\(=-1\)
b)\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}\)
\(=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0\)
\(=0\)
a.
\(\left(100+42\right)\times42+\left(200-58\right)\times58=142\times42+142\times58=142\times\left(42+58\right)=142\times100=14200\)
b.
\(\left(1998\div18-1443\div13\right)\times\left(16996-1110\div30\times305\right)=\left(111-111\right)\times\left(16996-1110\div30\times305\right)\)
\(=0\times\left(16996-1110\div30\times305\right)=0\)
c.
\(5^3_5+1,75+6^1_8+4^1_4+3,875+3,4=5,6+1,75+6,125+4,25+3,875+3,4\)
\(=\left(5,6+3,4\right)+\left(1,75+4,25\right)+\left(6,125+3,875\right)=9+6+10=25\)
Giải theo cách lớp 5
a)(100+42)x42+(200-58)x58
=142x42-142x58
=(42-58)x142
=(-16).142
=-2272
b)(1998:18-1443:13)x(16996-1110:30x305)
=(111-111)x(16996-1110:30x305)
=0x(16996-1110:30x305)
=0
c)\(5\frac{3}{5}\)+1,75+\(6\frac{1}{8}\)+\(4\frac{1}{4}\)+3,875+3,4
=(1,75+\(4\frac{1}{4}\))+(3,875+\(6\frac{1}{8}\))+(3,4+\(5\frac{3}{5}\))
=6+10+9
=25
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