1) \(\frac{4}{7}-\frac{1}{14}+\left|\frac{-5}{21}\right|\)

\(=\frac{4}{7}-\frac{1}{14}+\frac{5}{21}\)

\(=\frac{24}{42}-\frac{3}{42}+\frac{10}{42}\)

\(=\frac{31}{42}\)

2) \(\left|\frac{-2}{3}\right|-\frac{1}{2}+3\)

\(=\frac{2}{3}-\frac{1}{2}+\frac{3}{1}\)

\(=\frac{4}{6}-\frac{3}{6}+\frac{18}{6}\)

\(=\frac{19}{6}\)

3) \(\left|\frac{7}{-4}\right|-\frac{5}{8}+\frac{-2}{3}\)

\(=\frac{7}{4}-\frac{5}{8}+\frac{-2}{3}\)

\(=\frac{42}{24}-\frac{15}{24}+\frac{-16}{24}\)

\(=\frac{11}{24}\)

4) \(\frac{4}{5}+\left|\frac{-3}{2}\right|+\frac{1}{-4}\)

\(=\frac{4}{5}+\frac{3}{2}+\frac{-1}{4}\)

\(=\frac{16}{20}+\frac{30}{20}+\frac{-5}{20}\)

\(=\frac{41}{20}\)

5) \(\left|\frac{-1}{4}\right|-3+\frac{3}{4}\)

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

\(=\frac{1}{4}-\frac{12}{4}+\frac{3}{4}\)

\(=-2\)

6) \(\left|\frac{-1}{3}\right|-\frac{5}{4}+\frac{1}{5}\)

\(=\frac{1}{3}-\frac{5}{4}+\frac{1}{5}\)

\(=\frac{20}{60}-\frac{75}{60}+\frac{12}{60}\)

\(=\frac{-43}{60}\)

a, Ta có: \(\dfrac{1}{n}.\dfrac{1}{n+4}=\dfrac{1}{n.\left(n+4\right)}=\dfrac{1}{4}.\dfrac{4}{n.\left(n+1\right)}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)

Vậy \(\dfrac{1}{n}.\dfrac{1}{n+1}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)

b, \(A=\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+...+\dfrac{4}{95}.\dfrac{4}{99}=4.\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)\)

\(=4.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)\)

\(=4.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=4.\dfrac{32}{99}=\dfrac{128}{99}\)

Vậy \(A=\dfrac{128}{99}\)

a, C = 1 + 4 + 4² + 4³ + 4⁴ + 4⁵ + 4⁶

   4C = 4 + 4² + 4³ + 4⁴ + 4⁵ + 4⁶ + 4⁷

b, 4C - C = ( 4+4² + 4³ + 4⁴ +4⁵ + 4⁶ + 4⁷ ) - ( 1 + 4 + 4² + 4³ +4⁴ +4⁵ + 4⁶ )

3C = 4⁷ - 1

=> C = ( 4⁷ - 1 ) : 3

* Cách 1 : 

Ta có : 

\(16A=\frac{4^{17}+16}{4^{17}+1}=\frac{4^{17}+1+15}{4^{17}+1}=\frac{4^{17}+1}{4^{17}+1}+\frac{15}{4^{17}+1}=1+\frac{15}{4^{17}+1}\)

\(16B=\frac{4^{14}+16}{4^{14}+1}=\frac{4^{14}+1+15}{4^{14}+1}=\frac{4^{14}+1}{4^{14}+1}+\frac{15}{4^{14}+1}=1+\frac{15}{4^{14}+1}\)

Vì \(\frac{15}{4^{17}+1}< \frac{15}{4^{14}+1}\) nên \(1+\frac{15}{4^{17}+1}< 1+\frac{15}{4^{14}+1}\)

\(\Rightarrow\)\(16A< 16B\) hay \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

\(4^2.A=\frac{4^2\left(4^{15}+1\right)}{4^{17}+1}\); \(4^2.B=\frac{4^2\left(4^{12}+1\right)}{4^{14}+1}\)

=> \(4^2.A=\frac{4^{17}+4^2}{4^{17}+1}\);\(4^2.B=\frac{4^{14}+4^2}{4^{14}+1}\)

=> \(4^2.A=\frac{4^{17}+1+4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1+4^2-1}{4^{14}+1}\)

=> \(4^2.A=\frac{4^{17}+1}{4^{17}+1}+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1}{4^{14}+1}+\frac{4^2-1}{4^{14}+1}\)

=> \(4^2.A=1+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=1+\frac{4^2-1}{4^{14}+1}\)

Mà \(4^{17}>4^{14}\)

=> \(4^{17}+1>4^{14}+1\)

=> \(\frac{4^2-1}{4^{17}+1}< \frac{4^2-1}{4^{14}+1}\)

=> \(1+\frac{4^2-1}{4^{17}+1}< 1+\frac{4^2-1}{4^{14}+1}\)

=> \(4^2.A< 4^2.B\)

=> \(A< B\)

\(4A=4+4^2+...+4^{100}\)

\(A=1+4+4^2+..+4^{99}\)

\(\Rightarrow3A=4A-A=4^{100}-1\)

\(\Rightarrow3A< 4^{100}\)

\(\Rightarrow\frac{3A}{B}< 1\Rightarrow\frac{A}{B}< \frac{1}{3}\)

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