Tính hợp lí:
\(\dfrac{6^{36}.\left(50.5^{40}-10.5^{34}\right)}{30^{30}.10^4.\left(100.15^5-4.3^5\right)}\)
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\(=\dfrac{2^{36}.3^{36}\left(2.5^2.5^{40}-2.5.5^{34}\right)}{2^{30}.3^{30}.5^{30}.2^4.5^4\left(2^2.5^2.3^5.5^5-2^2.3^5\right)}=\dfrac{2^{36}.3^{36}\left(2.5^{42}-2.5^{35}\right)}{2^{34}.3^{30}.5^{34}\left(2^2.3^5.5^7-2^2.3^5\right)}\)
\(=\dfrac{2^2.3^6.2.5^{35}\left(5^7-1\right)}{2^2.3^5\left(5^7-1\right)}=3.2.5^{35}=6.5^{35}\)
120 - { 300: [ 40 - (2.562 - 262.5)] + 10}
= 120 - { 300 : [ 40 - (-186)] + 10}
= 120 - {300 : 226 + 10}
= 120 - \(\dfrac{1280}{113}\)
= \(\dfrac{12280}{113}\)
\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\dfrac{1}{6}-\dfrac{-10}{3}\)
\(=\dfrac{7}{2}\)
\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
Tính hợp lí:
a) \(\left(-0,4\right)+\dfrac{3}{8}+\left(-0,6\right)\)
\(=\left[\left(-0,4\right)+\left(-0,6\right)\right]+\dfrac{3}{8}\)
\(=-1+\dfrac{3}{8}\)
\(=\dfrac{\left(-8\right)+3}{8}\)
\(=\dfrac{-5}{8}\)
b) \(\dfrac{4}{5}-1,8+0,375+\dfrac{5}{8}\)
\(=\dfrac{4}{5}-\dfrac{9}{5}+\dfrac{3}{8}+\dfrac{5}{8}\)
\(=-1+1\)
\(=0\\\)
c) \(\dfrac{7}{3}.\left(-2,5\right).\dfrac{6}{7}\)
\(=\dfrac{7}{3}.\dfrac{-5}{2}.\dfrac{6}{7}\)
\(=\dfrac{7}{3}.\dfrac{6}{7}.\dfrac{-5}{2}\)
\(=2.\dfrac{-5}{2}\)
\(=-5\)
d) \(\dfrac{7}{12}.\left(-2,34\right)-\dfrac{7}{12}.\left(-0,34\right)\)
\(=\dfrac{7}{12}.\left[\left(-2,34\right)+0,34\right]\)
\(=\dfrac{7}{12}.\left(-2\right)\)
\(=\dfrac{-7}{6}\)
e) \(\dfrac{-8}{3}.\dfrac{2}{11}-\dfrac{8}{3}:\dfrac{11}{9}\)
\(=\dfrac{8}{3}.\dfrac{-2}{11}-\dfrac{8}{3}.\dfrac{9}{11}\)
\(=\dfrac{8}{3}.\left(\dfrac{-2}{11}-\dfrac{9}{11}\right)\)
\(=\dfrac{8}{3}.-1\)
\(=\dfrac{-8}{3}\)
Chúc bạn học tốt
a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)
c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)
d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)
\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)
\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)
e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)
\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)
\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)
f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)
\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)
\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)
g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)
\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)
\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)
\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)
\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)
\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)
\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)
\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)
\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)