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a,\(3\dfrac{17}{24}+\left(2\dfrac{8}{15}-4\dfrac{8}{15}\right):\left(2\dfrac{11}{30}-\dfrac{11}{30}\right)\)
\(=\dfrac{89}{24}-2:2\)
\(=\dfrac{65}{24}\)
b,\(0,5:\sqrt{625}-\sqrt{\dfrac{4}{25}}+0,18.\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right)\)
\(=0,5:25-\dfrac{2}{5}+0,18.\dfrac{1}{2}\)
\(=-\dfrac{29}{100}\)
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
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
\(A=\dfrac{\dfrac{3}{11}+\dfrac{3}{3}-\dfrac{3}{7}}{\dfrac{9}{11}+\dfrac{9}{3}-\dfrac{9}{7}}-\dfrac{\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{8}}{\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}+\dfrac{7}{16}}\)
\(=\dfrac{1}{3}-1:\dfrac{7}{2}=\dfrac{1}{3}-\dfrac{2}{7}=\dfrac{1}{21}\)
a: \(=\left(\dfrac{5^4}{5^2\cdot7^2}\right)^{15}:\left(\dfrac{5^6}{2\cdot7\cdot31}\right)^7\)
\(=\dfrac{5^{30}}{7^{30}}:\dfrac{5^{42}}{2^7\cdot7^7\cdot31^7}\)
\(=\dfrac{5^{30}}{7^{30}}\cdot\dfrac{2^7\cdot7^7\cdot31^7}{5^{42}}=\dfrac{2^7\cdot31^7}{7^{23}\cdot5^{12}}\)
b: \(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\)
\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-7\cdot4\right)}{5^{29}\cdot2^8\cdot7^{48}}=5\cdot\left(-27\right)=-135\)
\(\dfrac{9^{15}\cdot8^{11}}{3^{29}\cdot16^8}=\dfrac{3^{2^{15}}\cdot2^{3^{11}}}{3^{29}\cdot2^{4^8}}=\dfrac{3^{30}\cdot2^{33}}{3^{29}\cdot2^{32}}\)
\(\Rightarrow\dfrac{3^{30}\cdot2^{33}}{3^{29}\cdot2^{32}}=\dfrac{3^{29}\cdot2^{32}\cdot3\cdot2}{3^{29}\cdot2^{32}}=3\cdot2=6\)
16×3^10+120×6^9/4^6×3^12+6^11