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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)
\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)
b)
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)
a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)
\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)
\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)
\(=\left[9+1\right]+1+\dfrac{13}{17}\)
\(=10+1+\dfrac{13}{17}\)
\(=11+\dfrac{13}{17}\)
\(=\dfrac{187}{17}+\dfrac{13}{17}\)
\(=\dfrac{200}{17}\)
b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}+\dfrac{12}{7}\)
\(=\dfrac{7}{7}\)
\(=1\)
c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)
\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)
\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)
\(=\left[6+0\right]-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{42}{7}-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)
\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)
\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)
\(=\dfrac{2}{7}.\left[2+0\right]\)
\(=\dfrac{2}{7}.2\)
= \(\dfrac{4}{7}\)
1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)
a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)
=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)
=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)
=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)
b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)
=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))
=99+1=100
c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha
\(a,-13-\left(-15\right)=-13+15=2\)
\(b,43-\left(56-43\right)+\left(56-12\right)=43-13+44=30+44=74\)
\(c,4^4:2^8+\left[13-\left(-5\right)\right]\cdot\left(-3\right)=256:256+\left(13+5\right)\cdot\left(-3\right)\)\(=1+18\cdot\left(-3\right)=1+\left(-54\right)=-53\)
\(d,135\cdot\left(13-37\right)+37\cdot\left(135-13\right)=135\cdot\left(-24\right)+37\cdot122\)\(=-3240+4514=1274\)
a, -13-(-15)=-13+15=2
b, 43-(56-43)+(56-12)=43-13+44=30+44=74
c, 4^4:2^8+[13-(-5)].(-3)=256:256+(13+5).(-3)=1+18.(-3)=1+(-54)=-53
d, 135.(13-37)+37.(135-13)=135.(-24)+37.122=-3240+4514=1274