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Vì (x - 2)2 \(\ge\) 0 => (x - 2)2 - 15 \(\ge\) 0 - 15 = -15
(x - 2) - 15 \(\ge-15\)
a) Ta so sanh (x-2)2-15 va -15
Hay: (x-2)2 -15+15 va -15+15
Hay: (x-2)2 va 0
Ta thay: (x-2)2 lon hon hoac bang 0 nen suy ra:
(x-2)2-15 se lon hon hoac bang 15
b) Ta so sanh: 8/3 - |x+1/2| va 3
Hay: 8/3 - 8/3 - |x+1/2| va 3-8/3
Hay: - | x+1/2| va 1/3
Ta thay: |x+1/2| lon hon hoac bang 0 => -|x+1/2| se be hon hoac bang 0
=> - | x+1/2| < 1/3
=> 8/3 - |x+1/2| < 3
\(\frac{4}{9}=\frac{4.2}{9.2}=\frac{8}{18}\)
Vì : \(\frac{8}{18}<\frac{13}{18}\)
=> \(\frac{4}{9}<\frac{13}{18}\)
b) tương tự
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{99\cdot101}{100^2}\)
\(=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(=2\cdot101=202\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
a, \(\frac{1313}{2727}=\frac{13\cdot101}{27\cdot1001}=\frac{13}{27}\)
b,\(\frac{151515}{232323}=\frac{15.10101}{23.10101}=\frac{15}{23}\)
\(A=\frac{15}{1.6}+\frac{15}{6.11}+\frac{15}{11.16}+...+\frac{15}{2011.2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+..+\frac{5}{2011.2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{2011}-\frac{1}{2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=1-\frac{1}{2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=\frac{2015}{2016}\)
\(\Rightarrow\)\(A=\frac{2015}{672}\) (1)
Mà \(3=\frac{2016}{672}\) (2)
Từ (1) và (2) suy ra A < 3