fthực hiện phép tính :(2^3x9^4+9^3x45):(9^2x10-9^2)ai nhanh đúng thì mình tich cho 6sao
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\(\frac{4^5.9^5-2.6^{10}}{2^{10}.3^9+6^9.20}=\frac{\left(4.9\right)^5-2.2^{10.}.3^{10}}{2^{10}.3^9+2^9.3^9.2^2.5}\)
\(\frac{2^{10}.3^{10}^{ }-2^{11}.3^{10}}{2^{10}.3^9+2^{11}.3^9.5}=\frac{2^{10}.3^{10}\left(1-2\right)}{2^{10}.3^9\left(1+2.5\right)}\)
\(\frac{2^{10}.3^{10}.\left(-1\right)}{2^{10}.3^9.\left(-9\right)}=\frac{-3}{-9}=\frac{1}{3}\)
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
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
a] 4/12 ; 5/12 ; 11/2 ; 1/4
b] 1 ; 9/12; 12/5 ; 11/3
c] 8/21; 6/11;8/7
d] 2/3 ;2/7 15/2
a]1/3 5/12 11/2 1/4
b]1 3/4 12/5 33/9
c]8/21 6/11 8/7
d]2/3 2/7 15/2
\(\left(\frac{4}{3}-\frac{2}{3}-\frac{9}{8}\right):\left(1-\frac{4}{5}\right)\)
\(=\left(\frac{2}{3}-\frac{9}{8}\right):\left(1-\frac{4}{5}\right)\)
\(=-\frac{11}{24}:\frac{1}{5}\)
\(=-\frac{55}{24}\)