đặt A=100^10+1/100^10-1

B=10^100+1/10^100-3

ta có:\(A=\frac{100^{10}+1}{100^{10}-1}=\frac{100^{10}-1+2}{100^{10}-1}=\frac{100^{10}-1}{100^{10}-1}+\frac{2}{100^{10}-1}=1+\frac{2}{100^{10}-1}\)

\(B=\frac{10^{100}+1}{10^{100}-3}=\frac{10^{100}-3+4}{10^{100}-3}=\frac{10^{100}-3}{10^{100}-3}+\frac{4}{10^{100}-3}=1+\frac{4}{10^{100}-3}=1+\frac{4}{100^{10}-3}\)

vì 100¹⁰-1>100¹⁰-3

=>\(\frac{4}{100^{10}-1}<\frac{4}{100^{10}-3}\)

=>A<B

 Arcobaleno sai

Ta có:

\(\frac{100^{10}+1}{100^{10}-1}=\frac{100^{10}-1+2}{100^{10}-1}=\frac{100^{10}-1}{100^{10}-1}+\frac{2}{100^{10}-1}=1+\frac{2}{10^{20}-1}=1+\frac{4}{2.10^{20}-2}\)

\(\frac{10^{100}+1}{10^{100}-3}=\frac{10^{100}-3+4}{10^{100}-3}=\frac{10^{100}-3}{10^{100}-3}+\frac{4}{10^{100}-3}=1+\frac{4}{10^{100}-3}=1+\frac{4}{10^{100}-3}\)

Thấy: \(2.10^{20}-2<10^{100}-3\)

\(\Rightarrow\frac{4}{2.10^{20}-1}>\frac{4}{10^{100}-3}\)

\(\Rightarrow\frac{100^{10}+1}{100^{10}-1}>\frac{10^{100}+1}{10^{100}-3}\)

+> 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 \(B>A\)

hu hu ai trả lời giúp mình với 

>

<

=

\(\dfrac{7}{10}=\dfrac{70}{100}>\dfrac{43}{100};\dfrac{6}{10}=\dfrac{60}{100}< \dfrac{85}{100};\dfrac{7}{10}=\dfrac{70}{100}\)

\(\dfrac{7}{10}>\dfrac{43}{100};\dfrac{6}{10}< \dfrac{85}{100};\dfrac{7}{10}=\dfrac{70}{100}\)

a) Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};...;\frac{99}{100}< \frac{100}{101}\)

\(\Rightarrow A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

b) \(A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)

\(A.B=\frac{1.\left(3.5...99\right).\left(2.4.6...100\right)}{\left(2.4.6...100\right).\left(3.5.7...99\right).101}=\frac{1}{101}\)

c) vì A < b nên A . A < A . B < \(\frac{1}{101}< \frac{1}{100}\)

do đó : A . A  < \(\frac{1}{10}.\frac{1}{10}\)suy ra A < \(\frac{1}{10}\)

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,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)