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Ta có :
a = 123456.123458 = 123456 . ( 123457 + 1 ) = 123456 . 123457 + 123456
b = 123457 . 123457 = 123457 . ( 123456 + 1 ) = 123457 . 123456 + 123457
Mà 123456 < 123457 => 123456.123458 < 123457 . 123457 hay a < b
Vậy a < b
\(A=\frac{15^{16}+1}{15^{17}+1}\) và \(B=\frac{15^{15}+1}{15^{16}+1}\)
\(A< 1\Rightarrow A>\frac{15^{16}+1+14}{15^{17}+1+4}=\frac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}=B\)
\(\Rightarrow A< B\)
\(A=\frac{15^{16}+1}{15^{17}+1}=\frac{1}{225}\)
\(B=\frac{15^{15}+1}{15^{16}+1}=\frac{1}{225}\)
\(\Rightarrow A=B\)
=>A:1/2=1/1x3+1/3x5+1/5x7+...+1/99x101
=>2a=1/2(2/1x3+2/3x5+...+2/99x101)
từ đây tự làm
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)
\(\Rightarrow2A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(\Rightarrow4A=\frac{100}{101}\)
\(\Leftrightarrow A=\frac{100}{101}.\frac{1}{4}=\frac{4.25}{101.4}=25< 26\)
ta có : \(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(B=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(\Rightarrow\)\(B=A\)
456 x 128 / 451 x 128 =58368/57728
123 x 451 / 128 x 451 = 55473/57728
so sánh : 58368/57728 ...>.... 55473/ 57728
vậy suy ra : 456/451 ....>.... 123/128
tk mk nha mk nhanh nhất
\(\frac{456}{451}\) > \(\frac{123}{128}\)tích cho mik nhé
\(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{4}{5^5}+...+\frac{11}{5^{12}}\)
\(\Rightarrow\)\(5P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\)
\(\Rightarrow\)\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{1}{5^{12}}\)
\(\Rightarrow\)\(20P=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(\Rightarrow\)\(16P=1-\frac{1}{5^{11}}+\frac{1}{5^{12}}-\frac{1}{5^{11}}\)\(< 1\)
\(\Rightarrow\)\(P< \frac{1}{16}\)
P/s: nguyên tác: https://olm.vn/thanhvien/nhatphuonghocgiot
\(-\frac{123456}{123457}>-\frac{123457}{123455}\)