\(S=8+8^3+...+8^{2x+1}\\ \Rightarrow64S=8^3+8^5+...+8^{2x+3}\\ \Rightarrow64S-S=\left(8^3+8^5+...+8^{2x+3}\right)-\left(8+8^3+...+8^{2x+1}\right)\\ \Rightarrow63S=8^{2x+3}-8\\ \Rightarrow S=\dfrac{8^{2x+3}-8}{63}\)

Đề bảo tìm x thì thiếu đề r nhé

b: =>(x-4)(x-3)(x-1)>0

=>1<x<3 hoặc x>4

c: =>(2x-1)(x-1)(2x-3)<0

=>x<1/2 hoặc 1<x<3/2

1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)

\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)

=0

2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)

\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)

\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)

3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)

\(=\dfrac{52546}{15}\)

4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)

\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)

\(=\dfrac{4}{7}\)

cảm ơn nhé

a,C1 
 (5/6 + 5/8) x 2/3
= 35/24 x 2/3
=35/36
C2
 (5/6 + 5/8) x 2/3
= 5/6x2/3 + 5/8x2/3
= 10/18 + 10/24
= 35/36

b,C1
7/5 x 3/4 - 1/2 x 3/4
= 21/20 - 3/8
= 27/40
C2
7/5 x 3/4 - 1/2 x 3/4
=3/4 x (7/5-1/2)
=3/4x9/10
=27/40

a, 5/6 x (1/8x2/3) 

=5/6 x 1/12

=5/72

b,(7/5 - 1/2) x 3/4 

=0,9 x 3/4

=3,06

c,(5/6+5/8) x 2/3

=70/48 x 2/3

=35/36

a, 5/6 x (1/8x2/3) 

=5/6 x 1/12

=5/72

b,(7/5 - 1/2) x 3/4 

=0,9 x 3/4

=3,06

c,(5/6+5/8) x 2/3

=70/48 x 2/3

=35/36