(1+1+1+1+1+1+...) x X = (9+9++9+9+9+9+9+9+9) + 9
     100 số 1
 (1 x 100) x X = (9 x9) +9
100 x X = 81 +9
100 x X = 90
X = 90 : 100 
X = 0,9
Like nha!

(1+1+1+1+1+1+...)*x=(9+9+9+9+9+9+9+9+9)+9

 (100 số 1 ) 

(1×100)*x = (9×9)+9

100*x=81+9

100*x = 90

x= 90:100

x=0,9

(1+1+1+1+1+1+...) x X = (9+9++9+9+9+9+9+9+9) + 9
     100 số 1
 (1 x 100) x X = (9 x9) +9
100 x X = 81 +9
100 x X = 90
X = 90 : 100 

9*9*9*9*9*9*9*9*9*9*9*9+1*0+1*0+1

= 0+0+1

=1

9*9*9*9*9*9*9*9*9*9*9*9+1*0+1*0+1

=282429536481+1+1+1

=282429536484

9^12+1*0*2+1

\(\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{6}{18}+\dfrac{10}{18}=\dfrac{16}{18}=\dfrac{8}{9}\)

\(\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{9}{63}-\dfrac{7}{63}=\dfrac{2}{63}\)

\(3:\dfrac{5}{9}=3.\dfrac{9}{5}=\dfrac{27}{5}\)

\(3.\dfrac{5}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{1}{9}.\dfrac{9}{3}=\dfrac{1}{3}\)

\(\dfrac{1}{3}:\dfrac{1}{7}=\dfrac{7}{3}\)

\(9+\dfrac{9}{3}=9+3=12\)

\(4-\dfrac{2}{4}=4-\dfrac{1}{2}=\dfrac{7}{2}\)

\(\dfrac{1}{3}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{9}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3+5}{9}\) \(=\) \(\dfrac{8}{9}\)

\(\dfrac{1}{7}\) \(-\) \(\dfrac{1}{9}\) \(=\) \(\dfrac{9}{63}\) \(-\) \(\dfrac{7}{63}\) \(=\) \(\dfrac{9-7}{63}\) \(=\) \(\dfrac{2}{63}\)

\(\dfrac{1}{9}\) \(\times\) \(\dfrac{9}{3}\) \(=\) \(\dfrac{1\times9}{9\times3}\) \(=\) \(\dfrac{1}{3}\)

\(\dfrac{1}{3}\) \(\div\) \(\dfrac{1}{7}\) \(=\) \(\dfrac{1}{3}\) \(\times\) \(\dfrac{7}{1}\) \(=\) \(\dfrac{1\times7}{3\times1}\) \(=\) \(\dfrac{7}{3}\)

\(3\) \(\div\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{1}\) \(\div\)  \(\dfrac{5}{9}\) \(=\dfrac{3}{1}\times\dfrac{9}{5}=\dfrac{3\times9}{1\times5}=\dfrac{27}{5}\)

\(3\times\dfrac{5}{9}=\dfrac{3}{1}\times\dfrac{5}{9}=\dfrac{3\times5}{1\times9}=\dfrac{5}{3}\)

\(9+\dfrac{9}{3}=\dfrac{9}{1}+\dfrac{9}{3}=\dfrac{27}{3}+\dfrac{9}{3}=\dfrac{27+9}{3}=\dfrac{36}{3}=12\)

\(4\) \(-\dfrac{2}{4}=\dfrac{4}{1}-\dfrac{2}{4}=\dfrac{16}{4}-\dfrac{2}{4}=\dfrac{14}{4}=\dfrac{7}{2}\)

a) 9+9+9+9+9+9+9+9+9+9+9+9+9 = 9 x 13 = 117
b) 1+2+3+4+5+6+7+8+9 = (1+9)+(2+8)+(3+7)+(4+6)+5= 10 x 4 + 5+ = 40 + 5 = 45
c) 1+4+8+3+0+1+8+9+11+60=105 (bạn tự nghĩ nha chứ mình lười :v)
Học tốt :D

a, 117

b, 45

c, 105

bài của mình đây nha,  chúc bạn học tốt

a) kq=117

b) kq=46

c) kq=105

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Lời giải:

Sử dụng công thức $(a-1)(a+1)=a^2-1$ ta có:

$8F=(9-1)(9+1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$

$=(9^2-1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$

$=(9^4-1)(9^4+1)(9^8+1)...(9^{32}+1)$

$=(9^8-1)(9^8+1)...(9^{32}+1)$

$=(9^{16}-1)...(9^{32}+1)=(9^{32}-1)(9^{32}+1)=9^{64}-1$

$\Rightarrow F=\frac{9^{64}-1}{8}$