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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)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
\(\left(0.25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)
\(=0.4^{10}+\sqrt{25-9}\)
\(=0+\sqrt{16}=0+4=4\)
\(\dfrac{5}{20}+\dfrac{18}{11}-25\%-\left(\dfrac{18}{11}-\dfrac{4}{9}\right)\)
\(=\dfrac{5}{20}+\dfrac{18}{11}-\dfrac{1}{4}-\dfrac{18}{11}+\dfrac{4}{9}\)
\(=\left(\dfrac{5}{20}-\dfrac{1}{4}\right)+\left(\dfrac{18}{11}-\dfrac{18}{11}\right)+\dfrac{4}{9}\)
\(=0+0+\dfrac{4}{9}=\dfrac{4}{9}\)
a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)
b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)
c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)
d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
\(\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right):\dfrac{5}{9}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right):\dfrac{5}{9}\right]\cdot\dfrac{11}{235}\)
\(=\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right)\cdot\dfrac{9}{5}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{9}{5}\right]\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left(\dfrac{-3}{8}+\dfrac{11}{23}+\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left[\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{11}{23}+\dfrac{12}{23}\right)\right]\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left(-1+1\right)\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot0\cdot\dfrac{11}{235}\)
\(=0\)
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)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)
\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)
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) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)
c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)
a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)
\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)
\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)
\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)
\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)
b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\cdot\dfrac{4}{5}-1\)
\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)
\(\left(\dfrac{4}{9}-\dfrac{5}{11}\right):\dfrac{3}{10}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right):\dfrac{3}{10}-\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\left(-\dfrac{10}{3}\right)\\ =\left(\dfrac{4}{9}-\dfrac{5}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right)\cdot\dfrac{10}{3}-\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\left(-1\right)\cdot\dfrac{10}{3}\\ =\left(\dfrac{4}{9}-\dfrac{5}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\dfrac{10}{3}=\dfrac{10}{3}\cdot\left(\dfrac{4}{9}-\dfrac{5}{11}+\dfrac{3}{9}-\dfrac{9}{11}+\dfrac{2}{9}-\dfrac{8}{11}\right)\\ =\dfrac{10}{3}\cdot\left(-1\right)\\ =-\dfrac{10}{3}\)
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.