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*Trả lời :
a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
=\(\dfrac{3}{4}.\left(-8\right)\)
= \(-6\)
b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
=\(10:\left(-\dfrac{5}{7}\right)\)
=\(-14\)
c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )
=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)
=\(\left(-\dfrac{5}{5}+\dfrac{9}{9}\right):\dfrac{7}{11}\)
=\(\left(-1+1\right):\dfrac{7}{11}\)
\(=0:\dfrac{7}{11}\)
=0.
d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
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
Lời giải:
a)
\(\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}}{\frac{8}{3}-\frac{8}{5}+\frac{8}{7}-\frac{8}{9}+\frac{8}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}\) \(=\frac{2}{8}=\frac{1}{4}\)
b)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)
\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)
Ta có \(P_1>0,P_2< 0,P_3=0\) (Vì có thừa số \(\dfrac{0}{11}=0\))
Do đó \(P_2< P_3< P_1\)
\(\dfrac{4}{9}:\left(\dfrac{-1}{7}\right)+6\dfrac{5}{9}.\left(\dfrac{2}{3}\right)\)
\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}.\dfrac{2}{3}\)
\(=\dfrac{2}{9}.\left(-14\right)+\dfrac{2}{9}.\dfrac{59}{3}\)
\(=\dfrac{2}{9}.\left(-14+\dfrac{59}{3}\right)\)
\(=\dfrac{2}{9}.\dfrac{17}{3}\)
\(=\dfrac{34}{27}\)
\(\left(\dfrac{-1}{3}\right)^2.\dfrac{4}{11}+\dfrac{7}{11}.\left(\dfrac{-1}{3}\right)^2\)
\(=\dfrac{1}{9}.\dfrac{4}{11}+\dfrac{7}{11}.\dfrac{1}{9}\)
\(=\dfrac{1}{9}.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\)
\(=\dfrac{1}{9}.1=\dfrac{1}{9}\)
~ \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}.\dfrac{2}{3}\)
\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}.\dfrac{2}{3}\)
\(=-\dfrac{28}{9}+\dfrac{118}{27}\)
\(=-\dfrac{84}{27}+\dfrac{118}{27}\)
\(=\dfrac{34}{27}\)
~ \(\left(-\dfrac{1}{3}\right)^2.\dfrac{4}{11}+\dfrac{7}{11}.\left(-\dfrac{1}{3}\right)^2\)
\(=\left(-\dfrac{1}{3}\right)^2.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\)
\(=\dfrac{1}{9}.\dfrac{11}{11}\)
\(=\dfrac{1}{9}.1\)
\(=\dfrac{1}{9}\)
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
\(A=\dfrac{5}{9}.\dfrac{22}{-3}+\dfrac{5}{9}.\dfrac{5}{-3}=\dfrac{5}{9}\left(\dfrac{22}{-3}+\dfrac{5}{-3}\right)=\dfrac{5}{9}.\dfrac{22+5}{-3}=\dfrac{5}{9}.\dfrac{27}{-3}\)
\(A=\dfrac{5.27}{9.-3}=\dfrac{5.27}{-27}=-5\)
\(B=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(B=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)\(B=\left(6-5-3\right)+\left(\dfrac{-2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(B=-2+\left(\dfrac{-2-5+7}{2}\right)+\left(\dfrac{1+3-5}{2}\right)\)
\(B=-2+\dfrac{0}{2}+\left(\dfrac{-1}{2}\right)=-2+0-\dfrac{1}{2}=\dfrac{-5}{2}\)
\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)
\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)
\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)
\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)
\(=-\dfrac{304}{189}\)
\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)
\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)
\(=\dfrac{-22489}{7680}\)
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,
\(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\\ =\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:-\dfrac{3}{5}\\ =\dfrac{5}{9}:\left(\dfrac{-3}{22}+-\dfrac{3}{5}\right)\\ =\dfrac{5}{9}:-\dfrac{81}{110}\\ =-\dfrac{550}{729}\)
\(\left(\dfrac{4}{9}-\dfrac{5}{11}\right):\dfrac{3}{10}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right):\dfrac{3}{10}-\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\left(-\dfrac{10}{3}\right)\\ =\left(\dfrac{4}{9}-\dfrac{5}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right)\cdot\dfrac{10}{3}-\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\left(-1\right)\cdot\dfrac{10}{3}\\ =\left(\dfrac{4}{9}-\dfrac{5}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\dfrac{10}{3}=\dfrac{10}{3}\cdot\left(\dfrac{4}{9}-\dfrac{5}{11}+\dfrac{3}{9}-\dfrac{9}{11}+\dfrac{2}{9}-\dfrac{8}{11}\right)\\ =\dfrac{10}{3}\cdot\left(-1\right)\\ =-\dfrac{10}{3}\)
Ta có:(4/9-5/11):3/10+(3/9-9/11):3/10-(2/9-8/11).(-10/3)
=[(4/9-5/11)+(3/9-9/11)]:3/10+(-2/9+8/11).(-10/3)
=[(4/9+3/9)+(-5/11-9/11)]:3/10+(-2/9+8/11):(-3/10)
=(7/9-14/11):3/10+(2/9-8/11):3/10 (nhân chuyển dấu)
=[(7/9-14/11)+(2/9-8/11)]:3/10
=(1-2):3/10
=-1.10/3
=-10/3.