\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

\(B=\frac{7}{8^3}+\frac{3}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

\(A=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^3}>B=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

vậy A>B

A=(24+7)/8^4 =31/8^4 
B=(56+3)/8^4 = 59/8^4 
=> A<B

A = 48 . 12 = 576

B = 64 + 512 = 576

=> A = B

Vậy,..........

so sánh A = 48 . ( 4 + 8 ) và B = 4 mũ 3 + 8 mũ 3

          =   A = 48.12

          =   A = 576

          =B = 4^{3 +}8³

            =B=576

VÌ 576=576 NÊN A=B

\(8-\frac{3}{2\cdot4}+\frac{3}{4\cdot6}+...+\frac{3}{98\cdot10}\)

\(=8-\frac{3}{2}\left[\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{98\cdot100}\right]\)

\(=8-\frac{3}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right]\)

\(=8-\frac{3}{2}\left[\frac{1}{2}-\frac{1}{100}\right]=8-\frac{3}{2}\cdot\frac{49}{100}=8-\frac{147}{200}=\frac{1453}{200}>1\)

a: 8/9<1<9/3

b: 8/4=2=6/3

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=

a) 8/9 < 1 < 9/3

b) 8/4 = 2 = 6/3

hep me

\(\dfrac{3}{4}=\dfrac{3.3}{4.3}=\dfrac{9}{12}\)

Ta có :\(\dfrac{9}{12}>\dfrac{8}{12}\)

=>\(\dfrac{3}{4}>\dfrac{8}{12}\)

\(MSC:12\\ \dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\\ \dfrac{9}{12}>\dfrac{8}{12}\\ \Rightarrow\dfrac{3}{4}>\dfrac{8}{12}\)