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\(a,(\frac{1}{4}+\frac{-5}{13})+(\frac{2}{11}+\frac{-8}{13}+\frac{3}{4})\)
\(= (\frac{1}{4} + \frac{3}{4}) + (\frac{-5}{13} + \frac{-8}{13}) + \frac{2}{11}\)
\(= \frac{4}{4} + \frac{-13}{13} + \frac{2}{11}\)
\(=1+(-1)+\frac{2}{11}=\frac{2}{11}\)
\(b,(\frac{21}{31}+\frac{-16}{7})+(\frac{44}{53}+\frac{10}{31})+\frac{9}{53}\)
\(= (\frac{21}{31} + \frac{10}{31}) + (\frac{44}{53} + \frac{9}{53}) + \frac{-16}{7}\)
\(=\frac{31}{31}+\frac{53}{53}+\frac{-16}{7}=1+1-\frac{16}{7}\)
\(=2-\frac{16}{7}=\frac{14}{7}-\frac{16}{7}=-\frac27\)
\(c,\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
\(= (\frac{-5}{7} + \frac{-2}{7}) + (\frac{3}{4} + \frac{1}{4}) + \frac{-1}{5}\)
\(= \frac{-7}{7} + \frac{4}{4} + \frac{-1}{5}\)
\(=-1+1-\frac{1}{5}=0-\frac15=-\frac15\)
\(\frac{-3}{31} + \frac{-6}{17} + \frac{1}{25} + \frac{-28}{31} + \frac{-11}{17} + \frac{-1}{5}\)
\(= (\frac{-3}{31} + \frac{-28}{31}) + (\frac{-6}{17} + \frac{-11}{17}) + \frac{1}{25} + \frac{-1}{5}\)
\(= \frac{-31}{31} + \frac{-17}{17} + \frac{1}{25} - \frac{5}{25}\)
\(= -1 + (-1) + \frac{-4}{25}\)
\(=-2-\frac{4}{25}=-\frac{50}{25}-\frac{4}{25}=-\frac{54}{25}\)
a,\(=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
=-1+1+-2/11
=0+-2/11
=-2/11
b,\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{40}\right)+\frac{-5}{17}\)
=1+-1+-5/17
=0+-5/17
=-5/17
c,\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+-\frac{7}{9}\right)+\frac{16}{17}\)
=1+-1+16/17
=0+16/17
=16/17
d,\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
a.\(\frac{-5}{9}\)+\(\frac{8}{15}\)+\(\frac{-2}{11}\)+\(\frac{4}{-9}\)+\(\frac{7}{15}\)
=\(\frac{-5}{9}\)+\(\frac{4}{-9}\)+\(\frac{8}{15}\)+\(\frac{7}{15}\)+\(\frac{-2}{11}\)
=(\(\frac{-5}{9}\)+\(\frac{-4}{9}\))+(\(\frac{8}{15}\)+\(\frac{7}{15}\))+\(\frac{-2}{11}\)
=(-1)+1+\(\frac{-2}{11}\)
=0+\(\frac{-2}{11}\)
=\(\frac{-2}{11}\).
\(\frac27\times5\frac14-\frac27\times3\frac14\)
=\(\frac27\times\left(5\frac14-3\frac14\right)\)
=\(\frac27\times\left(\left(5-3\right)+\left(\frac14-\frac15\right)\right)\)
=\(\frac27\times\left(2+0\right)\)
=\(\frac27\times2\)
=\(\frac47\)
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
b)
c)
d)
e)
f)
g) (hoặc )
h) (hoặc )
i) (hoặc )
k)
l)
m)
n)
p)
q) Giải chi tiết từng bước 1. Nhóm các phân số cùng mẫu số a)
(Lưu ý: Đề bài ghi là , nếu đề là hoặc khác đi thì kết quả sẽ thay đổi, ở đây tính theo đề chữ). b) c) d) e) 2. Sử dụng tính chất phân phối f) g) h) k) m) 3. Các phép tính khác i) n) p) Chuyển hết sang phân số hoặc số thập phân để nhóm:
(Phép này nên nhóm phân số mẫu 4, 8, 12).
Kết quả: q) r) Kết quả các phép tính lần lượt là: a) ; b) ; c) ; d) ; e) ; f) ; g) ; h) ; i) ; k) ; m) ; n) ; p) ; q) ; r) .
wow
\(a,(\frac{3}{7}-\frac{4}{5})+4\cdot\frac{1}{5}=\frac{15 - 28}{35}+\frac{4}{5}\)
\(=\frac{-13}{35}+\frac{28}{35}=\frac{15}{35}=\frac{3}{7}\)
\(b,\frac{2}{-11}+0,25-\frac{9}{11}+\frac{3}{17}-\frac{3}{-4}=\frac{-2}{11}+\frac{1}{4}-\frac{9}{11}+\frac{3}{17}+\frac{3}{4}\)
\(=(\frac{-2}{11}-\frac{9}{11})+(\frac{1}{4}+\frac{3}{4})+\frac{3}{17}=\frac{-11}{11}+\frac{4}{4}+\frac{3}{17}\)
\(=-1+1+\frac{3}{17}=0+\frac{3}{17}=\frac{3}{17}\)
\(c,\frac{10}{17}-\frac{5}{13}-(\frac{-7}{17})-\frac{8}{13}+\frac{11}{25}\)
\(= \frac{10}{17} - \frac{5}{13} + \frac{7}{17} - \frac{8}{13} + \frac{11}{25}\)
\(= (\frac{10}{17} + \frac{7}{17}) - (\frac{5}{13} + \frac{8}{13}) + \frac{11}{25}\)
\(= \frac{17}{17} - \frac{13}{13} + \frac{11}{25}\)
\(=1-1+\frac{11}{25}=\frac{11}{25}\)
\(d,6\frac{2}{5}-(2\frac{4}{9}+4\frac{2}{5})=6\frac{2}{5}-2\frac{4}{9}-4\frac{2}{5}\)
\(=(6\frac{2}{5}-4\frac{2}{5})-2\frac{4}{9}=2-2\frac{4}{9}\)
\(=2-\frac{22}{9}=\frac{18 - 22}{9}=\frac{-4}{9}\)
\(e,(\frac{1}{4}+\frac{-5}{13})+(\frac{2}{11}+\frac{-8}{13}+\frac{3}{4})\)
\(= \frac{1}{4} - \frac{5}{13} + \frac{2}{11} - \frac{8}{13} + \frac{3}{4}\)
\(= (\frac{1}{4} + \frac{3}{4}) - (\frac{5}{13} + \frac{8}{13}) + \frac{2}{11}\)
\(=\frac{4}{4}-\frac{13}{13}+\frac{2}{11}=1-1+\frac{2}{11}=\frac{2}{11}\)
\(f,\frac{-4}{9}\cdot\frac{3}{11}+\frac{-4}{9}\cdot\frac{8}{11}+\frac{-5}{9}=\frac{-4}{9}\cdot(\frac{3}{11}+\frac{8}{11})-\frac{5}{9}\)
\(=\frac{-4}{9}\cdot\frac{11}{11}-\frac{5}{9}=\frac{-4}{9}\cdot1-\frac{5}{9}\)
\(=\frac{-4}{9}-\frac{5}{9}=\frac{-9}{9}=-1\)
\(g,\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{4}{7}-\frac{-3}{5}=\frac{-3}{5}\cdot(\frac{5}{7}+\frac{4}{7}-1)\)
\(=\frac{-3}{5}\cdot(\frac{9}{7}-\frac{7}{7})=\frac{-3}{5}\cdot\frac{2}{7}=\frac{-6}{35}\)
\(h,5\frac{2}{7}\cdot\frac{8}{11}+5\frac{2}{7}\cdot\frac{5}{11}-5\frac{2}{7}\cdot\frac{2}{11}=5\frac{2}{7}\cdot(\frac{8}{11}+\frac{5}{11}-\frac{2}{11})\)
\(=5\frac{2}{7}\cdot\frac{11}{11}=5\frac{2}{7}\cdot1=\frac{37}{7}\)
\(i,0,5\cdot1\frac{1}{3}\cdot0,75:\frac{2}{5}+\frac{3}{5}=\frac{1}{2}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{2}+\frac{3}{5}\)
\(=(\frac{1 \cdot4 \cdot3 \cdot5}{2 \cdot3 \cdot4 \cdot2})+\frac{3}{5}=\frac{5}{4}+\frac{3}{5}\)
\(=\frac{25 + 12}{20}=\frac{37}{20}\)
\(k,\frac{-5}{8}\cdot\frac{2}{13}+\frac{-5}{8}\cdot\frac{11}{13}+1=\frac{-5}{8}\cdot(\frac{2}{13}+\frac{11}{13})+1\)
\(=\frac{-5}{8}\cdot\frac{13}{13}+1=\frac{-5}{8}\cdot1+1\)
\(=\frac{-5}{8}+\frac{8}{8}=\frac{3}{8}\)
\(m,\frac{12}{15}\cdot\frac{-22}{47}+\frac{-25}{47}:\frac{15}{12}=\frac{4}{5}\cdot\frac{-22}{47}+\frac{-25}{47}\cdot\frac{12}{15}\)
\(=\frac{4}{5}\cdot\frac{-22}{47}+\frac{-25}{47}\cdot\frac{4}{5}=\frac{4}{5}\cdot(\frac{-22}{47}+\frac{-25}{47})\)
\(=\frac{4}{5}\cdot\frac{-47}{47}=\frac{4}{5}\cdot(-1)=-\frac{4}{5}\)
\(n,\frac{12}{7}\cdot\frac{7}{4}+\frac{35}{11}:\frac{245}{121}=(\frac{12}{4}\cdot\frac{7}{7})+\frac{35}{11}\cdot\frac{121}{245}\)
\(=3+(\frac{35}{245}\cdot\frac{121}{11})=3+(\frac{1}{7}\cdot11)\)
\(=3+\frac{11}{7}=\frac{21 + 11}{7}=\frac{32}{7}\)
\(p,4\frac{3}{4}+(-0,37)+\frac{1}{8}+(-1,28)+(-2,5)+3\frac{1}{12}\)
\(= \frac{19}{4} - \frac{37}{100} + \frac{1}{8} - \frac{128}{100} - \frac{250}{100} + \frac{37}{12}\)
\(= \frac{475}{100} - \frac{37}{100} - \frac{128}{100} - \frac{250}{100} + \frac{1}{8} + \frac{37}{12}\)
\(= \frac{60}{100} + \frac{1}{8} + \frac{37}{12}\)
\(=\frac{3}{5}+\frac{1}{8}+\frac{37}{12}=\frac{24 + 5}{40}+\frac{37}{12}\)
\(=\frac{29}{40}+\frac{37}{12}=\frac{87 + 370}{120}=\frac{457}{120}\)
\(q,(-1,7)\cdot(1-3,82)-3,82\cdot1,7\)
\(= -1,7 \cdot 1 + (-1,7) \cdot (-3,82) - 3,82 \cdot 1,7\)
\(= -1,7 + (1,7 \cdot 3,82) - (3,82 \cdot 1,7)\)
\(=-1,7+0=-1,7\)
\(r,(1,7-22,9)+(1,7-25+22,9)\)
\(= 1,7 - 22,9 + 1,7 - 25 + 22,9\)
\(= (1,7 + 1,7) + (-22,9 + 22,9) - 25\)
\(=3,4+0-25=-21,6\)
a) \(\left(\frac37-\frac45\right)+4.\frac15\)
\(=\frac37-\frac45+\frac45\)
\(=\frac37+0\)
\(=\frac37\)
b) \(\frac{2}{-11}+0,25-\frac{9}{11}+\frac{3}{17}-\frac{3}{-4}\)
\(=\left(-\frac{2}{11}-\frac{9}{11}\right)+\left(\frac14+\frac34\right)+\frac{3}{17}\)
\(=-1+1+\frac{3}{17}\)
\(=0+\frac{3}{17}\)
\(=\frac{3}{17}\)
c) \(\frac{10}{17}-\frac{5}{13}-\left(\frac{-7}{17}\right)-\frac{8}{13}+\frac{11}{25}\)
\(=\left(\frac{10}{17}+\frac{7}{17}\right)-\left(\frac{5}{13}+\frac{8}{13}\right)+\frac{11}{25}\)
\(=1-1+\frac{11}{25}\)
\(=\frac{11}{25}\)
d) \(6\frac25-\left(2\frac49+4\frac25\right)\)
\(=\frac{32}{5}-\frac{22}{9}-\frac{22}{5}\)
\(=\left(\frac{32}{5}-\frac{22}{5}\right)-\frac{22}{9}\)
\(=2-\frac{22}{9}\)
\(=-\frac49\)