Biểu thức A có chứa thừa số 169 -13.13 = 0

=> A = 0

\(A=1-1-\dfrac{5}{6}+1+\dfrac{7}{12}-1-\dfrac{9}{20}+1+\dfrac{11}{30}-1-\dfrac{13}{42}+1+\dfrac{15}{56}-1-\dfrac{17}{72}+1+\dfrac{19}{90}\)

\(=1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}+\dfrac{1}{10}\)

=1/2+1/10

=5/10+1/10=6/10=3/5

1/2+5/6+11/12+19/20+29/30+41/42+55/56

= 1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90

= 9-(1/2+1/6+1/12+1/20+1/42+1/58+1/72+1/90)

=9-( 1/1.2 + 1/2.3 +1/3.4 +1/4.5 +1/5.6+ 1/6.7 +1/7.8 +1/8.9 +1/9.10)

=9-(1 -1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)

=9-(1 -1/10)

=9-(9/10)= 81/10=8,1

đề mình chỉ đến 55/56 thôi bạn

a) 103 + 20 + 5 = 123 + 5 = 128

Giá trị của biểu thức

103 + 20 + 5 là 128.

b) 241 – 41 + 29 = 200 + 29 = 229

Giá trị của biểu thức

241 – 41 + 29 là 229.

c) 516 – 10 + 30 = 506 + 30 = 536

Giá trị của biểu thức

516 – 10 + 30 là 536.

d) 635 – 3 – 50 = 650 – 50 = 600

Giá trị của biểu thức

635 – 3 – 50 là 600.

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trong biểu thức có chứa(169-13.13)

mà (169-13.13)=0

=>A=0

Tính giá trị biểu thức:

A = (169 - 1 . 1) . (169 - 2 . 2) . ... . (169 - 23 . 23)????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????/

\(\sqrt{64}-\sqrt{169}+\sqrt{9}=8-13+3=-2\)

\(4\sqrt{3}+\sqrt{27}-\sqrt{75}=4\sqrt{3}+3-5\sqrt{3}=\sqrt{3}\left(4+3-5\right)=2\sqrt{3}\)

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)

\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(2A=3A-A\)

\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(2A=\frac{3}{4}-\frac{1}{927}\)

\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)

\(A=\frac{182}{243}:\frac{1}{2}\)

\(A=\frac{364}{243}\)

A=1/4+1/12+1/36+1/108+1/324+1/972

=243/972+81/972+27/972+9/972+3/972+1/972

=364/972

=91/243

\(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}\)

\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)

\(B=1.9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}\right)\)

\(B=9-\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{56}\right)\right]\)

\(B=9-\left(1-\frac{1}{56}\right)\)

\(B=9-\frac{55}{56}\)

\(B=\frac{449}{56}\)

a. 54 : 6 + 41                                                   b. 83 - 7 x 9

     =     9  + 41                                                     = 83 -  63     

               =        50                                                          =     20      

41 x 5 - 100 = 205 – 100 = 105

\(=\cot49^0-\cot49^0=0\)