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a: \(\dfrac{A}{B}=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-9\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\)
b: \(\dfrac{C}{D}=\dfrac{3^7\cdot727+3^{12}}{3^{11}+2^3\cdot3^{11}+723\cdot3^6}\)
\(=\dfrac{3^7\left(727+3^5\right)}{3^6\left(3^5+2^3\cdot3^5+723\right)}=\dfrac{1}{3}\cdot3=1\)
tạm thời những bài này bn hãy tính = máy tính , nếu tính bằng giấy sẽ rất mất thì h
\(5.4^{15}.9^9-4.3^{20}.8^9\)
\(=5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9\)
\(=5.2^{30}.3^{18}-2^2.3^{20}.2^{27}\)
\(=5.2^{29}.3.3^{18}-3^{18}.3^2.2^{29}\)
\(=3^{18}.2^{29}.\left(15-9\right)\)
\(=3^{18}.2^{29}.6\)
\(=3^{18}.2^{29}.2.3\)
\(=3^{19}.2^{30}\)
\(5.2^6.6^{19}-7.2^{29}.27^6\)
\(=5.2^6.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6\)
\(=5.2^6.2^{19}.3^{19}-7.2^{29}.3^{18}\)
\(=5.2^{25}.3^{18}.3-7.2^{25}.2^4.3^{18}\)
\(=2^{25}.3^{18}.\left(15-112\right)\)
\(=2^{25}.3^{18}.\left(-97\right)\)
5.415.99-4.320.89=5.230.318-229.320=229.318.(5.2-32)=229.318
5.29.619-7.229.276=5.228.319-7.229.318=228.318.(5.3-7.2)=228.318
Đề bài sai ak bn :
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6}\)
\(=\frac{5.2^{30}.3^{18}-3^{20}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}.\left(5.2-3^2\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}=\frac{2^{29}.3^{18}.1}{2^{28}.3^{18}.1}=\frac{2^{29}}{2^{28}}=2\)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)