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Ở câu a) số mũ lúc nào cug dương mà bạn ( 45-10 = 4510). Nếu số mũ là dương thì:
a)\(\frac{45^{10}.5^{20}}{75^{15}}\)
= \(\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}\)
= \(\frac{3^{20}.5^{10}.5^{20}}{3^{15}.5^{30}}\)
= \(\frac{3^{20}.5^{30}}{3^{15}.5^{30}}\)
= \(\frac{3^5.1}{1.1}\)
= \(\frac{243}{1}\)
= 243
b)\(\frac{2^{15}.9^4}{6^6.8^2}\)
= \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^2}\)
= \(\frac{2^{15}.3^8}{2^6.3^6.2^6}\)
= \(\frac{2^{15}.3^8}{2^{12}.3^6}\)
= \(\frac{2^3.3^2}{1.1}\)
= \(\frac{8.9}{1}\)
= \(\frac{72}{1}\)
= 72
Đặt \(A=1+2+2^2+2^3+...+2^{2008}\)
\(2A=2.\left(1+2+2^2+2^3+...+2^{2008}\right)\)
\(2A=2+2^2+2^3+...+2^{2009}\)\(2A-A=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)\)
\(A=2^{2009}-1\)
\(\Rightarrow S=\frac{2^{2009}-1}{1-2^{2009}}\)
\(S=\frac{2^{2009}-1}{-\left(-1+2^{2009}\right)}=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=-1\)
\(\frac{2^3.3}{2^2.3^2.5}=\frac{2}{3.5}=\frac{2}{15}\)
Thiếu dấu nhân ở chỗ \(2^2.3^2\)nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
\(A=\frac{2^{100}-1}{2^{100}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
hok tốt!!
A = 1 + 2 + 22 + 23+......+29
2A = 2+ 22 + 23 +.......+29 + 210
2A - A = 210 -1
A = 210 - 1
11+12+123=123553444