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a, \(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right)-\dfrac{5}{21}:\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\\ = -\dfrac{5}{21}:\dfrac{4}{5}+ \left(-\dfrac{5}{21}\right):\dfrac{4}{5}\\ =\left[-\dfrac{5}{21}+\left(-\dfrac{5}{21}\right)\right]:\dfrac{4}{5}\\ -\dfrac{10}{21}:\dfrac{4}{5}\\ =-\dfrac{25}{42}\)
b,
\(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\\ =\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:-\dfrac{3}{5}\\ =\dfrac{5}{9}:\left(\dfrac{-3}{22}+-\dfrac{3}{5}\right)\\ =\dfrac{5}{9}:-\dfrac{81}{110}\\ =-\dfrac{550}{729}\)
\(B=\dfrac{4}{9}:\left(\dfrac{1}{5}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\)
\(B=\dfrac{4}{9}:\dfrac{-7}{15}+\dfrac{4}{9}:\dfrac{-3}{22}\)
\(B=\dfrac{4}{9}.\dfrac{15}{-7}+\dfrac{4}{9}.\dfrac{22}{-3}\)
\(B=\dfrac{4}{9}.\left(\dfrac{15}{-7}+\dfrac{22}{-3}\right)\)
\(B=\dfrac{4}{9}.\dfrac{-199}{21}\)
\(B=\dfrac{-796}{189}\)
\(A=\dfrac{5}{9}.\dfrac{22}{-3}+\dfrac{5}{9}.\dfrac{5}{-3}=\dfrac{5}{9}\left(\dfrac{22}{-3}+\dfrac{5}{-3}\right)=\dfrac{5}{9}.\dfrac{22+5}{-3}=\dfrac{5}{9}.\dfrac{27}{-3}\)
\(A=\dfrac{5.27}{9.-3}=\dfrac{5.27}{-27}=-5\)
\(B=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(B=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)\(B=\left(6-5-3\right)+\left(\dfrac{-2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(B=-2+\left(\dfrac{-2-5+7}{2}\right)+\left(\dfrac{1+3-5}{2}\right)\)
\(B=-2+\dfrac{0}{2}+\left(\dfrac{-1}{2}\right)=-2+0-\dfrac{1}{2}=\dfrac{-5}{2}\)
\(B=\dfrac{\dfrac{2}{10}-\dfrac{3}{8}+\dfrac{5}{11}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{-2}{3}-\dfrac{1}{3}=-1\)
6)a) \(\left|\dfrac{5}{3}:x\right|=\left|\dfrac{-1}{6}\right|\)
⇒ \(\left|\dfrac{5}{3}:x\right|=\dfrac{1}{6}\)
⇒ \(\dfrac{5}{3}:x=\dfrac{1}{6}\) hoặc \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)
*TH1 : \(\dfrac{5}{3}:x=\dfrac{1}{6}\)
⇒ \(x=\dfrac{5}{3}:\dfrac{1}{6}=10\)
*TH2 : \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)
⇒ \(x=\dfrac{5}{3}:\dfrac{-1}{6}=-10\)
Vậy \(x\) ∈ \(\left\{10;-10\right\}\)
\(b,\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|\dfrac{-3}{4}\right|\)
⇒ \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
⇒\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
⇒ \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\) hoặc \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)
TH1 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\)
⇒ \(\dfrac{3}{4}x=\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)
⇒\(x=\dfrac{9}{4}:\dfrac{3}{4}=3\)
TH2 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)
⇒ \(\dfrac{3}{4}x=\dfrac{-3}{2}+\dfrac{3}{4}=\dfrac{-3}{4}\)
⇒ \(x=\dfrac{-3}{4}:\dfrac{3}{4}=-1\)
Vậy \(x\) ∈ \(\left\{3;1\right\}\)
a: \(=\dfrac{11}{21}+\dfrac{10}{21}+\dfrac{-2}{7}+\dfrac{-5}{7}-\dfrac{6}{5}=\dfrac{-6}{5}\)
b: \(=\left[0.25\cdot\left(-4\right)\right]^5\cdot\left(-\dfrac{50}{9}\right)\)
=50/9
c: \(=\dfrac{3}{22}\left(19+\dfrac{1}{7}+2+\dfrac{6}{7}\right)-6\)
\(=\dfrac{3}{22}\cdot22-6=3-6=-3\)
đề hồi nãy thì dấu chia còn đề này thì nhân
rốt cuộc là nhân hay chia
1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4
=1/4:1/4-2.-1/8+2
= 1-(-1/4)+2
=1+1/4+2=13/4
2) 3-(-6/7)^0+căn 9 :2
= 3-1+3:2
=3-1+3/2=7/2
3) (-2)^3+1/2:1/8-căn 25 + |-64|
= -8+4-5+64= 55
4) (-1/2)^4+|-2/3|-2007^0
= 1/16+2/3-1
= -13/48
5) = 178/495:623/495-17/60:119/120
= 2/7-2/7=0
6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]
= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]
= -1/2+[1+1+1/2]
= -1/2+5/2=2
Mấy cái dấu chấm đó là nhân nha bn!
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
\(\dfrac{5}{9}:(\dfrac{1}{11}-\dfrac{5}{22})+\dfrac{5}{9}:(\dfrac{1}{15}-\dfrac{2}{3})\)
\(=\dfrac{5}{9}:(\dfrac{2}{22}+\dfrac{-5}{22})+\dfrac{5}{9}:(\dfrac{1}{15}+\dfrac{-10}{15})\)
\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\dfrac{-9}{15}\)
\(=\dfrac{5}{9}\times\dfrac{-22}{3}+\dfrac{5}{9}\times\dfrac{-15}{9}\)
=\(\dfrac{5}{9}\times\left(-\dfrac{22}{3}+\dfrac{-15}{9}\right)\)
\(=\dfrac{5}{9}\times\left(-9\right)\)
\(=\left(-5\right)\)
\(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)=\dfrac{5}{9}:\left(\dfrac{-3}{22}\right)+\dfrac{5}{9}:\left(\dfrac{-3}{5}\right)\)
\(=\dfrac{5}{9}.\dfrac{22}{-3}+\dfrac{5}{9}.\dfrac{5}{-3}=\dfrac{5}{9}\left(\dfrac{22}{-3}+\dfrac{5}{-3}\right)=\dfrac{5}{9}.\dfrac{22+5}{-3}=\dfrac{5}{9}.\dfrac{27}{-3}\)
\(=\dfrac{5.27}{9.-3}=\dfrac{5.27}{-27}=-5\)