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\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}.\left(5.2-3^2\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}\)
\(=\frac{2^{29}.3^{18}.1}{2^{28}.3^{18}.1}\)
\(=2\)
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.2^{19}.3^{19}-7.2^{29}.\left(3^3\right)^6}\)
\(=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{29}.2.3^{18}-2^{29}.3^{18}.3^2}{5.2^{28}.3^{18}.3-7.2^{28}.2.3^{18}}\)
\(=\frac{2^{29}.3^{18}.\left(5.2-3^2\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}\)
\(=\frac{2^{29}.3^{18}.\left(10-9\right)}{2^{28}.3^{18}.\left(15-14\right)}\)
\(=\frac{2^{29}.3^{18}}{2^{28}.3^{18}}\)
\(=\frac{2^{28}.2.3^{18}}{2^{28}.3^{18}}\)
\(=2\)
a) \(\left(x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\left(x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(x+\frac{1}{3}=\frac{-2}{3}\)
\(x=-1\)
b) \(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\frac{25}{9}\)
\(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\left(\frac{5}{3}\right)^2\)
\(\frac{1}{3}x+\frac{4}{3}=\frac{5}{3}\)
\(\frac{1}{3}x=\frac{1}{3}\)
\(x=1\)
c) \(2^x+2^{x+1}=24\)
\(2^x+2^x.2=24\)
\(2^x.\left(1+2\right)=24\)
\(2^x.3=24\)
\(2^x=8\)
\(2^x=2^3\)
\(x=3\)
a, (x+1/3)^3 = -8/27
=>(x+1/3)^3 = (-2/3)^3
=>x+1/3 = -2/3
=>x = -1
b, (1/3x+4/3)^2 = 25/9
=>(1/3x+4/3)^2 = (5/3)^2
=>(1/3x+4/3) = 5/3
=>1/3x = 1/3
=> x = 1
c, 2^x + 2^x+1 = 24
=>2^x + 2^x . 2 = 24
=>2^x.(1+2) = 24
=>2^x . 3 = 24
=>2^x =8
=>2^x = 2^3
=> x = 3
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.3^{19}.2^{19}-7.2^{29}.\left(3^3\right)^6}=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{3^{18}.\left(5.2^{30}-2^{29}.3^2\right)}{3^{18}.\left(5.2^{28}.3-7.2^{29}\right)}=\frac{2^{28}.\left(5.2\right)}{2^{28}\left(5.3-7.2\right)}=\frac{10}{15-14}=\frac{10}{1}=10\)
Có j sai mong bỏ qua
Học tốt
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{2}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)(đáp án của p sai nha)
= 1 / 1*2 + 1 / 2*3 + 1/ 3*4 + 1 / 4 * 5 ... + 1/ 9*10 = 1-1/2 + 1/2 -1/3 +... + 1/9 - 1/10 = 1 - 1/10 = 9 /10
đáp án của bạn bị sai rùi
Bài giải
\(N=\frac{9\cdot5^{20}\cdot27^9-3\cdot9^{15}\cdot25^9}{7\cdot3^{29}\cdot125^6-3\cdot3^9\cdot15^{19}}=\frac{3^2\cdot5^{20}\cdot\left(3^3\right)^9-3\cdot\left(3^2\right)^{15}\cdot\left(5^2\right)^9}{7\cdot3^{29}\cdot\left(5^3\right)^6-3\cdot3^9\cdot\left(3\cdot5\right)^{19}}=\frac{3^2\cdot5^{20}\cdot3^{27}-3\cdot3^{30}\cdot5^{18}}{7\cdot3^{29}\cdot5^{18}-3^{10}\cdot3^{19}\cdot5^{19}}\)
\(=\frac{3^{29}\cdot5^{20}-3^{31}\cdot5^{18}}{7\cdot3^{29}\cdot5^{18}-3^{29}\cdot5^{19}}=\frac{3^{29}\cdot5^{18}\left(5^2-3^2\right)}{3^{29}\cdot5^{18}\left(7-5\right)}=\frac{25-9}{7-5}=\frac{16}{2}=8\)