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Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
\(C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\)
\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{21}\)
\(C=\dfrac{2}{7}\)
9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)
10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)
\(B=\dfrac{9}{10!}+\dfrac{10}{11!}+...........+\dfrac{99}{100!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{10}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
..........................
\(\dfrac{99}{100!}< \dfrac{100-1}{100!}=\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+...........+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
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
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
+> Ta đi chứng minh tính chất \(\frac{a}{b}>1\)thì \(\frac{a}{b}>\frac{a+c}{b+c}\)
Có\(\frac{a}{b}>1\Rightarrow a>b\)
\(\Rightarrow ac>bc\) \(\Rightarrow ac+ab>bc+ab\)\(\Rightarrow a\left(b+c\right)>b\left(a+c\right)\)\(\Rightarrow\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(1\right)\)
+> Aps dụng tính chất (1) vào b thức B ta có:
\(B=\frac{100^{10}-1}{100^{10}-3}>\frac{100^{10}-1+2}{100^{10}-3+2}=\frac{100^{10}+1}{100^{10}-1}\)
\(\Rightarrow B>\frac{100^{10}+1}{100^{10}-1}\)
\(\Rightarrow B>A\)
Vậy \(B>A\)
M=\(\dfrac{10^{100^{ }}+1}{10^{101}+1}\)
M=\(\dfrac{10^{99+1}+1}{10^{100+1}+1}\)
M=\(\dfrac{10^{99}.10+1}{10^{100}.10+1}\)
N=\(\dfrac{10^{99^{ }}+1}{10^{100}+1}\)
=>M lớn hơn N
M>N,vì:\(\dfrac{10^{100}+1}{10^{101}+1}=\dfrac{10^{100}}{10^{101}}\)
\(\dfrac{10^{99}+1}{10^{100}+1}=\dfrac{10^{99}}{10^{100}}\)
\(\dfrac{10^{100}}{10^{101}}>\dfrac{10^{99}}{10^{100}}\)
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