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Gọi \(\dfrac{12}{23}+\dfrac{12}{2323}-\dfrac{121212}{232323}\) là A
Ta sẽ tính biểu thức A.\(\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)=A.\(\left(\dfrac{7}{12}-\dfrac{7}{12}\right)=0\)
Vậy \(\left(\dfrac{12}{23}+\dfrac{12}{2323}-\dfrac{121212}{232323}\right).\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)=0
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước ko có tâm trạng mà cười nữa
1) \(19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12}\)
\(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\\ =\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)\\ =\dfrac{12}{7}\cdot\dfrac{35}{8}\\ =\dfrac{15}{2}\)
2) \(\dfrac{2}{5}\cdot\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}\cdot5\\ =\dfrac{1}{3}\cdot1-\dfrac{2}{3}\\ =\dfrac{1}{3}-\dfrac{2}{3}\\ =-\dfrac{1}{3}\)
3) \(\dfrac{4}{9}\cdot19\dfrac{1}{3}-\dfrac{4}{9}\cdot39\dfrac{1}{3}\)
\(=\dfrac{4}{9}\left(19\dfrac{1}{3}-39\dfrac{1}{3}\right)\\ =\dfrac{4}{9}\cdot\left(\dfrac{58}{3}-\dfrac{118}{3}\right)\\ =\dfrac{4}{9}\cdot\left(-20\right)\\ =-\dfrac{80}{9}\)
+) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)
Vậy \(A=\dfrac{2^{10}-1}{2^{10}}\)
+) \(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)
\(\Rightarrow\dfrac{1}{2}F=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{380}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{5}-\dfrac{1}{20}=\dfrac{3}{20}\Rightarrow F=\dfrac{3}{20}:\dfrac{1}{2}=\dfrac{3}{10}\)
Vậy \(F=\dfrac{3}{10}\)
+) \(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)
\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}=\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{25.28}\)
\(=\dfrac{4}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)
Vậy \(G=\dfrac{2}{7}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(A=1-\dfrac{1}{2^{10}}=\dfrac{1024-1}{1024}=\dfrac{1023}{1024}\)
\(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)
\(=\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{380}\)
\(=\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{19.20}\)
\(=2\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=2.\dfrac{3}{20}=\dfrac{3}{10}\)
\(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)
\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}\)
\(=\dfrac{4}{4.7}+\dfrac{4}{7.10}+\dfrac{4}{10.13}+...+\dfrac{4}{25.28}\)
\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
a) \(\dfrac{1}{3}+\dfrac{3}{8}-\dfrac{7}{12}\)
\(=\dfrac{17}{24}-\dfrac{7}{12}\)
\(=\dfrac{1}{8}\)
b) \(\dfrac{-3}{14}+\dfrac{5}{8}-\dfrac{1}{2}\)
\(=\dfrac{23}{56}-\dfrac{1}{2}\)
\(=\dfrac{-5}{56}\)
c) \(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{11}{18}\)
\(=\dfrac{-5}{12}-\dfrac{11}{18}\)
\(=\dfrac{-37}{36}\)
d) \(\dfrac{1}{4}+\dfrac{5}{12}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{2}{3}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{23}{39}-\dfrac{7}{8}\)
\(=\dfrac{-89}{312}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(-\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(-\dfrac{1}{12}+\dfrac{1}{12}\right)+\left(-\dfrac{1}{13}+\dfrac{1}{13}\right)+\left(-\dfrac{1}{14}+\dfrac{1}{14}\right)+\left(-\dfrac{1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\\ =\dfrac{1}{16}\)
Tính nhanh :
\(\dfrac{1}{10}+\dfrac{-1}{11}+\dfrac{1}{12}+\dfrac{-1}{13}+\dfrac{1}{14}+\dfrac{-1}{15}+\dfrac{1}{16}+\dfrac{-1}{10}+\dfrac{1}{11}+\dfrac{-1}{12}+\dfrac{1}{13}+\dfrac{-1}{14}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(\dfrac{-1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{12}+\dfrac{-1}{12}\right)+\left(\dfrac{-1}{13}+\dfrac{1}{13}\right)+\left(\dfrac{1}{14}+\dfrac{-1}{14}\right)\)
\(+\left(\dfrac{-1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\)
\(=0+0+...+0+\dfrac{1}{16}\)
\(=\dfrac{1}{16}\)
a) \(=\dfrac{-5}{12}.\dfrac{2}{7}+\dfrac{7}{12}.-\dfrac{3}{14}\)
\(=-\dfrac{5}{42}-\dfrac{1}{8}\)
= \(-\dfrac{41}{168}\)
b) \(=\dfrac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}\)
\(=\dfrac{1.1.3.\left(-2\right)}{\left(-1\right).2.1.\left(-2\right)}=\dfrac{3}{\left(-1\right).2}\)
\(=-\dfrac{3}{2}\)
\(12-12\dfrac{1}{3}=12-\dfrac{12\times3+1}{3}=12-\dfrac{37}{3}=\dfrac{12\times3-37}{3}=\dfrac{36-37}{3}=\dfrac{-1}{3}\)
= 12 và 2/3\(\dfrac{ }{ }\)