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\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
Tính A=1+1/2+1/3+1/4+...+1/2^100-1 rồi so sánh với 100
Làm ơn làm ơn giúp mk T_T ...
Nhanh mk tick cho
Ta có A = \(\frac{100^9+4}{100^9-1}=\frac{100^9-1+5}{100^9-1}=1+\frac{5}{100^9-1}\)
B = \(\frac{100^9+1}{100^9-4}=\frac{100^9-4+5}{100^9-4}=1+\frac{5}{100^9-4}\)
Vì \(\frac{5}{100^9-1}>\frac{5}{100^9-4}\Rightarrow1+\frac{5}{100^9-1}>1+\frac{5}{100^9-4}\Rightarrow A>B\)
\(A=\dfrac{2021^{10}-2021+2020}{2021^9-1}\\ =\dfrac{2021\left(2021^9-1\right)+2020}{2021^9-1}\\ =2021+\dfrac{2020}{2021^9-1}\\ B=\dfrac{2021^{11}-1}{2021^{10}-1}=2021+\dfrac{2020}{2021^{10}-1}\)
Ta có:
\(2021^9-1< 2021^{10}-1\\ \Rightarrow\dfrac{2020}{2021^9-1}>\dfrac{2020}{2021^{10}-1}\)
Do đó A > B.
\(B=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\)
\(B=\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}\)
\(B=-\left(\frac{3}{4}.\frac{8}{9}...\frac{99}{100}\right)\)
\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{9.11}{10.10}\right)\)
\(B=-\left(\frac{1.2...9}{2.3...10}.\frac{3.4...11}{2.3...10}\right)\)
\(B=-\left(\frac{1}{10}.\frac{11}{2}\right)\)
\(B=\frac{-11}{20}< \frac{-11}{21}\)
Vậy \(B< \frac{-11}{21}\)
Ta có:
\(78^{98}>78^{88}\)
\(\Leftrightarrow78^{98}\left(78-1\right)>78^{88}\left(78-1\right)\)
\(\Leftrightarrow78^{99}-78^{98}>78^{89}-78^{88}\)
\(\Leftrightarrow78^{99}+78^{88}>78^{89}+78^{98}\)
\(\Leftrightarrow78^{187}+78^{99}+78^{88}+1>78^{187}+78^{98}+78^{89}+1\)
\(\Leftrightarrow\left(78^{99}+1\right)\left(78^{88}+1\right)>\left(78^{98}+1\right)\left(78^{89}+1\right)\)
\(\Leftrightarrow\frac{78^{99}+1}{78^{89}+1}>\frac{78^{98}+1}{78^{88}+1}\).
Vậy \(A>B\).
cảm ơn bạn