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a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
\(A=\dfrac{10^{99}+1}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10\left(10^{99}+1\right)}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{100}+10}{10^{100}+1}=\dfrac{10^{100}+1+9}{10^{100}+1}=1+\dfrac{9}{10^{100}+1}\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10\left(10^{100}+1\right)}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{101}+10}{10^{101}+1}=\dfrac{10^{101}+1+9}{10^{101}+1}=1+\dfrac{9}{10^{101}+1}\)
Do \(\dfrac{9}{10^{100}+1}>\dfrac{9}{10^{101}+1}\) nên \(10A>10B\)
\(\Rightarrow A>B\)
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)
\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)
\(B< \dfrac{10^{100}+10}{10^{101}+10}\)
\(B< \dfrac{10\left(10^{99}+1\right)}{10\left(10^{100}+1\right)}\)
\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)
\(B< A\)
Ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(\Leftrightarrow A< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\dfrac{10^{10}+1}{10^{11}+1}=B\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
Vậy...
Ta có :
\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)
mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)
\(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)
có : `1-1/(11^99)<1`
\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)
hay `B<1/10`
Ta có:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)
\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)
\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)
\(A=-\dfrac{1}{10}\)
Mà: \(10>9\)
\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)
\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)
\(\Rightarrow A>-\dfrac{1}{9}\)
Áp dụng tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ta có :
\(B=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{98}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}=A\)
\(\Leftrightarrow B>A\)
Ta áp dụng tính chất :
\(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ta có:
\(B=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{89}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}=A\)
\(\Leftrightarrow B>A\)
Chúc bạn học tốt!