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\(\dfrac{8^{14}}{4^4.64^5}=\dfrac{\left(2^3\right)^{14}}{\left(2^2\right)^4.\left(2^5\right)^5}=\dfrac{2^{42}}{2^8.2^{25}}=2^{42-\left(8+25\right)}=2^9\)
\(\dfrac{9^{10}.27^7}{81^7.3^{15}}=\dfrac{\left(3^2\right)^{10}.\left(3^3\right)^7}{\left(3^4\right)^7.3^{15}}=\dfrac{3^{20}.3^{21}}{3^{28}.3^{15}}=\dfrac{3^{20+21}}{3^{28+15}}=\dfrac{3^{41}}{3^{41}.3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)
, bạn nào làm đc mik tick cho
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
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
A=33=27
\(B=\dfrac{2^{11}.9^2}{3^5.2^8}=\dfrac{2^{11}.3^4}{3^5.2^8}=\dfrac{2^3}{3}=\dfrac{8}{3}\)
a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...
\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....
\(C=\dfrac{2^6\cdot3^{10}}{3^9\cdot2^6}=3\\ D=\dfrac{3^{24}\cdot3^{10}}{3^{21}\cdot3^{11}}=\dfrac{3^{34}}{3^{32}}=3^2=9\\ F=\dfrac{2^{45}\cdot5^{14}}{5^{15}\cdot2^{47}}=\dfrac{1}{2^2\cdot5}=\dfrac{1}{20}\\ G=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot5^4}=\dfrac{1\cdot5}{2}=\dfrac{5}{2}\)
C=3
D=9
F=1/20
G=5/2
Em ko giải chi tiết vì nó lâu
Mong thông cảm!