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a/ \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
Mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
\(\Leftrightarrow x+101=0\)
\(\Leftrightarrow x=-101\)
Vậy...
b/ Đặt :
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+.........+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+....+\dfrac{10^2-9^2}{9^2.10^2}\)
\(=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+....+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{10^2}< 1\)
\(\Leftrightarrow A< 1\left(đpcm\right)\)
Vậy...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x-10\right|+\left|x-5\right|=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A\ge\left|x-10+5-x\right|\)
\(\Leftrightarrow A\ge5\)
Dấu "=" xảy ra
\(\Leftrightarrow\left(x-10\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge10\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le10\\5\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\5\le x\le10\end{matrix}\right.\)
Vậy..
T làm biếng lắm; làm C thôi
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)
Làm tương tự ta được A > 1/15
câu a
\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)
\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)
a, \(A=1+2+2^2+2^3+..........+2^{49}+2^{50}\)
\(\Leftrightarrow2A=2+2^2+..............+2^{50}+2^{51}\)
\(\Leftrightarrow2A-A=\left(2+2^2+.........+2^{51}\right)-\left(1+2+......+2^{50}\right)\)
\(\Leftrightarrow A=2^{51}-1\)
a) \(A=1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\)
\(\Rightarrow2A=2\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(2A=2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\)
\(\Rightarrow2A-A=A=\left(2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\right)-\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(A=2^{51}-1\) vậy \(A=2^{51}-1\)
b) \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^5+...+\left(\dfrac{1}{2}\right)^{99}+\left(\dfrac{1}{2}\right)^{100}\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(2B=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+\dfrac{2}{2^4}+\dfrac{2}{2^5}+...+\dfrac{2}{2^{99}}+\dfrac{2}{2^{100}}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B-B=B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(B=1-\dfrac{1}{2^{100}}\) vậy \(B=1-\dfrac{1}{2^{100}}\)
a, \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ 3B=3+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\\ 3B-B=\left(3+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\\2B=3-\dfrac{1}{3^{2005}}\\ B=\dfrac{3-\dfrac{1}{3^{2005}}}{2}\)
b,
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\\ 5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\\ 5A-A=\left(5+5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}\right)\\ 4A=5^{51}-1\\ A=\dfrac{5^{51}-1}{4}\)
c,
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2-1}\right)......\left(\dfrac{1}{100^2-1}\right)\\ A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)......\left(\dfrac{1}{10000}-1\right)\\ A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\cdot\cdot\cdot\dfrac{9999}{10000}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\cdot\cdot\cdot\dfrac{99\cdot101}{100\cdot100}\\ A=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\\ A=\dfrac{1}{100}\cdot\dfrac{101}{2}\\ A=\dfrac{101}{200}\)
d,
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\\ A=\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}+2^{97}+...+2^1\right)\)
Đặt \(A=B-C\)
\(\Rightarrow B=\left(2^{100}+2^{98}+...+2^2\right)vàC=\left(2^{99}+2^{97}+...+2^1\right)\)
\(B=2^{100}+2^{98}+...+2^2\\ 4B=2^{102}+2^{100}+...+2^4\\ 4B-B=\left(2^{102}+2^{100}+...+2^4\right)-\left(2^{100}+2^{98}+...+2^2\right)\\ 3B=2^{102}-2^2\\ B=\dfrac{2^{102}-2^2}{3}\left(1\right)\)
\(C=2^{99}+2^{97}+...+2^1\\ 4C=2^{101}+2^{99}+...+2^3\\ 4C-C=\left(2^{101}+2^{99}+...+2^3\right)-\left(2^{99}+2^{97}+...+2\right)\\ 3C=2^{101}-2\\ C=\dfrac{2^{101}-2}{3}\left(2\right)\)
Từ (1) và (2) ta có :
\(A=\dfrac{2^{102}-2^2}{3}-\dfrac{2^{101}-2}{3}\\ A=\dfrac{2^{102}-2^2-2^{101}+2}{3}\\ A=\dfrac{2^{102}-2^{101}+2}{3}\)
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)