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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)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
\(=\frac{9^{15}.6^{30}}{27^{21}.8^{11}}\)
\(=\frac{\left(3^2\right)^{15}.3^{30}.2^{30}}{\left(3^3\right)^{21}.\left(2^3\right)^{11}}\)
\(=\frac{3^{30}.3^{30}.2^{30}}{3^{63}.2^{33}}\)
\(=\frac{3^{60}.2^{30}}{2^{63}.3^{33}}=\frac{1}{2^3.3^3}=\frac{1}{216}\)
làm mẫu một bài còn lại tương tự nha bn =)
\(\frac{9^{15}.6^{30}}{27^{21}.8^{11}}=\frac{\left(3^2\right)^{15}.\left(2.3\right)^{30}}{\left(3^3\right)^{21}.\left(2^3\right)^{11}}=\frac{3^{60}.2^{30}}{3^{63}.2^{33}}=\frac{1}{3^3.2^3}\)
\(\frac{45^{12}.49^7}{35^{13}.27^8}=\frac{\left(5.3^2\right)^{12}.\left(7^2\right)^7}{\left(5.7\right)^{13}.\left(3^3\right)^8}=\frac{5^{12}.3^{24}.7^{14}}{5^{13}.7^{13}.3^{24}}=\frac{7}{5}\)
#
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
a. = 1/20 + 5 - 1/2
= 101/20 - 1/2
= 91/20
b. = ( 6/15 - 3/5) - ( 7/8 + 2/16) + 3
= -1/5 - 1 + 3
= 9/5
c. = 15/7 . ( 3/5 - 8/5)
= 15/7 . ( -1)
= - 15/7
e. = -14/9 - 3/9
= -17/9
f. = 19/21 . ( 15/17 + 2/17) + 13/21
= 19/21 . 1 + 13/21
= 32/21
g. = 43/12 : 2 + 5/24
= 43/24 + 5/24
= 2
\(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
\(a)\dfrac{5}{4}.\dfrac{-12}{7}=\dfrac{5.\left(-12\right)}{4.7}=\dfrac{-60}{28}=\dfrac{-15}{7}\)
\(b)\dfrac{-4}{3}:\dfrac{13}{9}=\dfrac{-4}{3}.\dfrac{9}{13}=\dfrac{\left(-4\right).9}{3.13}=\dfrac{-36}{39}=\dfrac{-12}{13}\)
\(c)\dfrac{-5}{7}.\dfrac{49}{3}:\dfrac{7}{-6}=\dfrac{-5}{7}.\dfrac{49}{3}:\dfrac{-7}{6}=\dfrac{-35}{3}.\dfrac{-6}{7}=10\)
\(d)\left(-\dfrac{9}{25}\right):6=\dfrac{-3}{50}\)
Chúc bn học tốt!
a) \(=\left(\left(-\frac{1}{4}-\frac{5}{3}\right)+\frac{7}{33}\right)-\left(-\frac{15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\left(-\frac{23}{12}+\frac{7}{33}\right)+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(-\frac{23}{12}+\frac{15}{12}\right)+\left(\frac{9}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=-\frac{2}{3}-\frac{3}{11}+\frac{48}{49}\)
\(=\frac{65}{1617}\)
b) \(=\frac{11}{125}+\left(-\frac{17}{18}+\frac{4}{9}\right)+\left(-\frac{5}{7}+\frac{17}{14}\right)\)
\(=\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}\)
\(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{9}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
\(A=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{\left(7.7\right)^2}}{\dfrac{8}{2}-\dfrac{4}{9}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(A=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{2401}}{\dfrac{8}{2}-\dfrac{4}{9}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(A=\dfrac{\dfrac{6}{7}+\dfrac{1}{49}-\dfrac{1}{2401}}{\dfrac{32}{9}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(A=\dfrac{\dfrac{43}{49}-\dfrac{1}{2401}}{\dfrac{1604}{441}-\dfrac{4}{343}}\)
\(A=\dfrac{\dfrac{2106}{2401}}{3,625526401}\)
\(A=\dfrac{2106}{2401}:3,625526401\)
\(A=\dfrac{9477}{39172}\)
\(\dfrac{49^5+49^7+49^9}{7^{11}+7^{13}+7^{15}+7^{17}+7^{19}+7^{21}}\)
\(=\dfrac{7^{10}+7^{14}+7^{18}}{7^{11}\left(1+7^2\right)+7^{15}\left(1+7^2\right)+7^{19}\left(1+7^2\right)}\)
\(=\dfrac{7^{10}\left(1+7^4+7^8\right)}{7^{11}\left(1+7^2\right)\left(1+7^4+7^8\right)}=\dfrac{1}{7\left(1+7^2\right)}=\dfrac{1}{7\cdot50}=\dfrac{1}{350}\)