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Gọi \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)là \(S\)
\(S=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\\ S>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{100\cdot101}\\ S>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ S>\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{5}\)
Vậy \(S>\dfrac{1}{5}\)(đpcm)
a) \(\dfrac{18.34-18.124}{-36.17+9.\left(-4\right).13}=\dfrac{18.\left(34-124\right)}{-36.\left(17+13\right)}=\dfrac{18.\left(-90\right)}{36.\left(-30\right)}=\dfrac{1.\left(-3\right)}{2.\cdot\left(-1\right)}=\dfrac{3}{2}\)
b) \(\dfrac{393939-10101}{505050-70707}=\dfrac{10101.39-10101}{10101.50-10101.7}=\dfrac{10101.\left(39-1\right)}{10101.\left(50-7\right)}=\dfrac{38}{43}\)
a) \(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}=\dfrac{36.17+36.\left(-62\right)}{36.\left(-17\right)+35.\left(-13\right)}=\dfrac{36\left[17+\left(-62\right)\right]}{36\left[\left(-17\right)+\left(-13\right)\right]}=\dfrac{45}{20}=\dfrac{9}{4}\)
b) \(\dfrac{3.7.13.37.39-10101}{505050-70707}=\dfrac{10101.39-10101}{10101.50-10101.7}=\dfrac{10101\left(39-1\right)}{10101\left(50-7\right)}=\dfrac{38}{43}\)
\(\left(2^{19}.27^3+15.4^9.9^4\right):\left(6^9.2^{10}+12^{10}\right)\)
\(=\left[2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4\right]:\left[2^9.3^9.2^{10}+2^{10}.6^{10}\right]\)
\(=\left(2^{19}.3^9+3.5.2^{18}.3^8\right):\left(2^{19}.3^9+2^{10}.2^{10}.3^{10}\right)\)
\(=\left(2^{19}.3^9+5.3^9.2^{18}\right):\left(2^{19}.3^9+2^{20}.3^{10}\right)\)
\(=2^{18}.3^9.\left(1.2+5\right):2^{19}.3^9.\left(1+2.3\right)\)
\(=\left(2^{18}.3^9.7\right):\left(2^{18}.2.3^9.7\right)\)
\(=1:2\)
\(=0.5\)
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
a, Ta có: \(\dfrac{32}{37}>\dfrac{32}{54}>\dfrac{19}{54}\Rightarrow\dfrac{32}{37}>\dfrac{19}{54}\)
b, Ta có: \(\dfrac{18}{53}>\dfrac{18}{54}=\dfrac{1}{3}\Rightarrow\dfrac{18}{53}>\dfrac{1}{3}\left(1\right)\)
\(\dfrac{26}{78}=\dfrac{1}{3}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{18}{53}>\dfrac{26}{78}\)
c, Ta thấy: \(\dfrac{25}{103}< \dfrac{25}{100}=\dfrac{1}{4}\left(1\right)\)
\(\dfrac{74}{295}>\dfrac{74}{296}=\dfrac{1}{4}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{25}{103}< \dfrac{74}{295}\)
a) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}=\dfrac{2.3\left(1+2.2+7.7\right)}{3.5\left(1+2.2+7.7\right)}=\dfrac{2}{5}\)
b) \(\dfrac{-1997.1996+1}{\left(-1995\right)\left(-1997\right)+1996}=\dfrac{-1997\left(1995+1\right)+1}{1995.1997+1996}=\dfrac{-1997.1995+\left(-1997\right)+1}{1995.1997+1996}=\dfrac{-1997.1995+\left(-1996\right)}{1995.1997+1996}=-1\)
thanks nhé