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Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
\(C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\)
\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{21}\)
\(C=\dfrac{2}{7}\)
mọi người thật là nhẫn tâm
chẳng ai giúp mk
TRỜI ƠI!!! AI MS LÀ BN BÈ THỰC SỰ
Ko cs đứa mô trả lời chứ chi
Loại bn bè vs mấy ng chỉ là giả tạo thôi
\(B=\dfrac{9}{10!}+\dfrac{10}{11!}+...........+\dfrac{99}{100!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{10}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
..........................
\(\dfrac{99}{100!}< \dfrac{100-1}{100!}=\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+...........+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
mk giải cho câu A rồi tự suy mấy câu khác nhé!
ta có : A = 10^8 + 2/10^8 - 1
=> A = 10^8 - 1 + 3/10^8 - 1
=> A = 1+ 3/10^8 - 1
B = 10^8/10^8 - 3
=> B = 10^8 - 3 + 3/10^8 - 3
=> B = 1+ 3/10^8 - 3
vì 3/10^8 - 1 < 3/10^8 - 3
=> 1 + 3/10^8 - 1 < 1 + 3/10^8 - 3
=> A < B
vậy A < B
cách này cô dạy mk đó
ta có:\(A=\frac{100^{10}+1}{100^{10}-1}=\frac{100^{10}-1+2}{100^{10}-1}=\frac{100^{10}-1}{100^{100}-1}+\frac{2}{100^{10}-1}=1+\frac{2}{100^{10}-1}\)
\(B=\frac{100^{10}-1}{100^{10}-3}=\frac{100^{10}-3+2}{100^{10}-3}=\frac{100^{10}-3}{100^{10}-3}+\frac{2}{100^{10}-3}=1+\frac{2}{100^{10}-3}\)
vì 10010-1>10010-3
\(\Rightarrow\frac{2}{100^{10}-1}<\frac{2}{100^{10}-3}\)
=>A<B
a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{8}{9}\right).\left(-\dfrac{15}{16}\right)...\left(-\dfrac{99}{100}\right)\)
\(A=\dfrac{\left(-1\right).3}{2^2}.\dfrac{\left(-2\right).4}{3^2}.\dfrac{\left(-3\right).5}{4^2}....\dfrac{\left(-9\right).11}{10^2}\)
\(A=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-9\right)}{2.3.4....10}.\dfrac{3.4.5....11}{2.3.4....10}\)
\(A=\dfrac{-1}{10}.\dfrac{11}{2}=-\dfrac{11}{20}\)
Câu B tương tự nha bạn!!!
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}......\dfrac{-99}{100}\)
\(A=\dfrac{-1.3}{2.2}.\dfrac{-2.4}{3.3}.\dfrac{-3.5}{4.4}.....\dfrac{-9.11}{10.10}\)
\(A=\dfrac{-1.3.-2.4.-3.5.....-9.11}{2.2.3.3.4.4.....10.10}\)
\(A=\dfrac{-1.-2.-3......-9}{2.3.4......10}.\dfrac{3.4.5....11}{2.3.4...10}\)
\(A=\dfrac{-1}{10}.\dfrac{11}{2}=\dfrac{-11}{20}\)
\(B=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).....\left(\dfrac{1}{10}-1\right)\)
\(B=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}........\dfrac{-9}{10}\)
\(B=\dfrac{-1.-2.-3......-9}{2.3.4......10}\)
\(B=\dfrac{-1}{10}\)
Ta có :
\(A=\dfrac{100^{10}+1}{100^{10}-1}=\dfrac{100^{10}-1+2}{100^{10}-1}=\dfrac{100^{10}-1}{100^{10}-1}+\dfrac{2}{100^{10}-1}=1+\dfrac{2}{100^{10}-1}\)
\(B=\dfrac{100^{10}-1}{100^{10}-3}=\dfrac{100^{10}-3+2}{100^{10}-3}=\dfrac{100^{10}-3}{100^{10}-3}+\dfrac{2}{100^{10}-3}=1+\dfrac{2}{100^{10}-3}\)
\(\) Vì \(1+\dfrac{2}{100^{10}-1}< 1+\dfrac{2}{100^{10}-3}\Rightarrow A< B\)
good