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a: =11/7(-3/7+4/11-4/7+7/11)=0
b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)
a,
Đặt A = \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\right)\)
\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{95}+...+\dfrac{1}{3}-1\right)\)
2A= \(2.\left(\dfrac{1}{99}-1\right)\)
\(\Rightarrow\) A = \(\dfrac{1}{99}-1=\dfrac{-98}{99}\)
b, \(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
= \(\dfrac{3.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}{5.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}\)
= \(\dfrac{3}{5}+\dfrac{2}{5}=\dfrac{5}{5}=1\)
Chúc bn hc tốt <3
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
a) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)b)
\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{7}{7}.33\dfrac{1}{3}=\dfrac{7}{3}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{7}{3}.\left(-14\right)=-\dfrac{1}{6}\)
c,
\(\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{5}{7}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-5}{7}\right)=\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{7}{5}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-7}{5}\right)\)
\(\left(-\dfrac{7}{5}\right)\left(15\dfrac{1}{4}+2010-25\dfrac{1}{4}-2016\right)=\left(-\dfrac{7}{5}\right)\left(-10-6\right)=22,4\)
d,
\(\left(2017-\dfrac{3}{7}+\dfrac{9}{11}\right)-\left(2016-\dfrac{3}{7}+\dfrac{8}{17}\right)-\left(2015+\dfrac{9}{11}-\dfrac{8}{17}\right)=2017-\dfrac{3}{7}+\dfrac{9}{11}-2016+\dfrac{3}{7}-\dfrac{8}{17}-2015-\dfrac{9}{11}+\dfrac{8}{17}\)\(\left(2017-2016-2015\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{8}{17}+\dfrac{8}{17}\right)=-2014\)
Bạn ơi cho mình hỏi tại sao đề bài câu c là -5/7 mà bn lm -7/5
\(\dfrac{4}{9}:\left(\dfrac{-1}{7}\right)+6\dfrac{5}{9}.\left(\dfrac{2}{3}\right)\)
\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}.\dfrac{2}{3}\)
\(=\dfrac{2}{9}.\left(-14\right)+\dfrac{2}{9}.\dfrac{59}{3}\)
\(=\dfrac{2}{9}.\left(-14+\dfrac{59}{3}\right)\)
\(=\dfrac{2}{9}.\dfrac{17}{3}\)
\(=\dfrac{34}{27}\)
\(\left(\dfrac{-1}{3}\right)^2.\dfrac{4}{11}+\dfrac{7}{11}.\left(\dfrac{-1}{3}\right)^2\)
\(=\dfrac{1}{9}.\dfrac{4}{11}+\dfrac{7}{11}.\dfrac{1}{9}\)
\(=\dfrac{1}{9}.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\)
\(=\dfrac{1}{9}.1=\dfrac{1}{9}\)
~ \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}.\dfrac{2}{3}\)
\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}.\dfrac{2}{3}\)
\(=-\dfrac{28}{9}+\dfrac{118}{27}\)
\(=-\dfrac{84}{27}+\dfrac{118}{27}\)
\(=\dfrac{34}{27}\)
~ \(\left(-\dfrac{1}{3}\right)^2.\dfrac{4}{11}+\dfrac{7}{11}.\left(-\dfrac{1}{3}\right)^2\)
\(=\left(-\dfrac{1}{3}\right)^2.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\)
\(=\dfrac{1}{9}.\dfrac{11}{11}\)
\(=\dfrac{1}{9}.1\)
\(=\dfrac{1}{9}\)
\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-\dfrac{1}{95\cdot93}-...-\dfrac{1}{3\cdot1}\)
\(B=-\left(\dfrac{1}{3\cdot1}+\dfrac{1}{5\cdot3}+...+\dfrac{1}{97\cdot99}\right)\)
\(2B=-\left(\dfrac{2}{3\cdot1}+\dfrac{2}{5\cdot3}+...+\dfrac{2}{99\cdot97}\right)\)
\(2B=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(2B=-\left(1-\dfrac{1}{99}\right)\)
\(2B=-\dfrac{98}{99}\)
\(B=-\dfrac{98}{198}\)
Cậu ơi, \(\dfrac{1}{99\cdot97}\) là dương mà sao lại đưa vào ngoặc âm tất cả vậy nhỉ?
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
a) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{3}{4}\right)^2.\left(-1\right)^5\)
\(=\dfrac{8}{27}-\dfrac{9}{16}.\left(-1\right)\)
\(=\dfrac{8}{27}-\left(-\dfrac{9}{16}\right)\)
\(=\dfrac{8}{27}+\dfrac{9}{16}\)
\(=\dfrac{128}{432}+\dfrac{243}{432}\)
\(=\dfrac{371}{432}\)
b) \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(\dfrac{-1}{12}\right)^2\)
\(=12:\dfrac{1}{144}\)
\(=12.144\)
\(=1728\)
c) \(\dfrac{7}{22}:\dfrac{3}{11}+\dfrac{7}{22}:\dfrac{4}{11}\)
\(=\dfrac{7}{22}:\left(\dfrac{3}{11}+\dfrac{4}{11}\right)\)
\(=\dfrac{7}{22}:\dfrac{7}{11}\)
\(=\dfrac{7}{22}.\dfrac{11}{7}\)
\(=\dfrac{1}{2}\)
d) \(\dfrac{12}{35}.\left(\dfrac{7}{4}+\dfrac{13}{4}\right)-\dfrac{1}{3}\)
\(=\dfrac{12}{35}.5-\dfrac{1}{3}\)
\(=\dfrac{12}{7}-\dfrac{1}{3}\)
\(=\dfrac{36}{21}-\dfrac{7}{21}\)
\(=\dfrac{29}{21}\)
Lời giải:
a)
\(=\left(\frac{-3}{7}+\frac{4}{11}+\frac{-4}{7}+\frac{7}{11}\right):\frac{7}{11}=\left(\frac{-3-4}{7}+\frac{4+7}{11}\right):\frac{7}{11}=(-1+1):\frac{7}{11}=0\)
b)
Đặt biểu thức là $A$
\(-2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}-\frac{2}{97.99}\)
\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{97-95}{95.97}-\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}-\frac{2}{97.99}\)
\(=1-\frac{1}{97}-\frac{2}{97.99}=\frac{96.99-2}{97.99}\)
\(\Rightarrow A=\frac{1-48.99}{97.99}\)