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dễ mà bạn đây là bài cơ bản lớp 6 dấy
câu a nhé bạn bạn nếu ko làm kiểu khó thì đổi về phaan số bình thường nà sau đó tính trong ngoặc trước rồi tính xoong bỏ dấu ngoặc nhưng ko đổi dấu né thế lad đc tương tự như các câu dưới
a)\(8\frac{2}{3}:2\frac{1}{6}-2\frac{27}{51}=\frac{26}{3}.\frac{6}{13}-\frac{43}{17}=4-\frac{43}{17}=\frac{25}{17}\)
b)\(\frac{27}{20}.\frac{15}{4}+\frac{19}{8}=\frac{119}{16}\)
c)\(\left(\frac{1}{12}+\frac{5}{6}\right)+\left(\frac{13}{35}+\frac{23}{35}\right)=\frac{11}{12}+\frac{36}{35}=\frac{817}{420}\)
d)\(\frac{24}{37}.\left(\frac{13}{18}+\frac{2}{9}+\frac{1}{18}\right)=\frac{24}{37}.1=\frac{24}{37}\)
a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)
=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)
=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)
=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)
=\(1+0+14+1-\frac{19}{5}\)
=\(15+1-\frac{19}{5}\)
=\(16-\frac{19}{5}\)
=\(\frac{80}{5}-\frac{19}{5}\)
=\(\frac{61}{5}\)
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)\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}\)
\(=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0\)
\(=0\)
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)\(5\frac14+3\frac25-4\frac14\)
=\(\left(5\frac14-4\frac14\right)+3\frac25\)
=\(\left\lbrack\left(5-4\right)+\left(\frac14-\frac14\right)\right\rbrack+\frac{17}{5}\)
=\(1+0+\frac{17}{5}\)
=\(\frac55+\frac{17}{5}\)
=\(\frac{22}{5}\)
13/50+9/100+41/100+12/50
=(13/50+12/50)+(9/100+41/100)
=1/2+1/2
=1
11) Ta có:
\(\frac{120-0,5.40.5.0,2.20.0,25-20}{1+5+9+...+33+37}\)
\(=\frac{120-\left(0,5.40\right).\left(5.0,2\right).\left(20.0,25\right)-20}{1+5+9+...+33+37}\)
\(=\frac{120-20.1.5-20}{1+5+9+...+33+37}\)
\(=\frac{120-100-20}{1+5+9+...+33+37}\)
\(=\frac{0}{1+5+9+...+33+37}=0\)
bài 1 a)
\(\frac{15}{7}\times\frac43+\frac63\)
=\(\frac{60}{21}+\frac63\)
=\(\frac{60}{21}+\frac{42}{21}\)
=\(\frac{102}{21}\)
=\(\frac{34}{7}\)
b)\(\frac{-13}{41}\times\frac{17}{11}\times\frac{11}{17}\)
=\(\frac{-13}{41}\times\left(\frac{17}{11}\times\frac{11}{17}\right)\)
=\(\frac{-13}{41}\times1\)
=\(\frac{-13}{41}\)
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 \(x=\frac{1}{13}\)