a; A = |-101| + |21| + |-99|  - |25|

   A = 101 + 21  + 99 - 25

 A = (101 + 99) - (25 - 21)

A = 200  -  4

A = 196 

b; B = ||17 - 42|  - 64|

    B = ||-25| - 64|

   B =  |25 - 64|

   B =  |-39|

  B = 39

c, C = |2⁷ - 7²| + |3³ - 3⁴| + |10³ - 3⁵|

   C = |128 - 49| + |27 - 81| + |1000 -  243|

   C = |79| + |-54| + | 757|

   C = 79 + 54 + 757

   C = 133 + 757

  C = 890 

d; D = |\(\dfrac{1}{3}\) - \(\dfrac{2}{3}\)| + |\(\dfrac{5}{6}\) + (-\(\dfrac{7}{12}\))|

    D = |-\(\dfrac{1}{3}\)|  + |\(\dfrac{1}{4}\)|

   D = \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)

   D  = \(\dfrac{7}{12}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}< 1\)

Vậy A<1 (điều cần chứng minh)

THANKS

not thing

\(A=\dfrac{4}{3x5}+\dfrac{4}{5x7}+\dfrac{4}{7x9}+...+\dfrac{4}{97x99}+\dfrac{4}{99x101}\)

\(A=4x\left(\dfrac{1}{3x5}+\dfrac{1}{5x7}+\dfrac{1}{7x9}+...+\dfrac{1}{97x99}+\dfrac{1}{99x101}\right)\)

\(A=4x\left[\dfrac{1}{2}x\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}x\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\dfrac{1}{2}x\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\dfrac{1}{2}x\left(\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}x\left(\dfrac{1}{99}-\dfrac{1}{101}\right)\right]\)

\(A=4x\dfrac{1}{2}x\left[\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right]\)

\(A=2x\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2x\dfrac{98}{303}=\dfrac{916}{303}\)

Giúp mình nhé

4/3x5+4/5x7+...+4/99x101=2(1/3-1/5+1/5-1/7+...+1/99-1/101)

=2(1/3-1/101)=2x98/303=196/303.

\(A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)\cdot3^5+\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}\right)\cdot3^9+...+\left(\frac{1}{3^{97}}+\frac{1}{3^{98}}+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\cdot3^{101}\)=\(\left(\frac{3^5}{3}+\frac{3^5}{3^2}+\frac{3^5}{3^3}+\frac{3^5}{3^4}\right)+\left(\frac{3^9}{3^5}+\frac{3^9}{3^6}+\frac{3^9}{3^7}+\frac{3^9}{3^8}\right)+...+\left(\frac{3^{101}}{3^{97}}+\frac{3^{101}}{3^{98}}+\frac{3^{101}}{3^{99}}+\frac{3^{101}}{3^{100}}\right)\)

=(3+3²+3³+3⁴)+(3+3²+3³+3⁴)+...+(3+3²+3³+3⁴)

Tổng trên có số số hạng là(mỗi ngoặc là 1 số hạng)

(101-5):4+1=25(số hạng)

=>A=25.(3+3²+3³+3⁴)=25.120=3000

Hướng dẫn giải:

a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)