làm lại

ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)

=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)

=>\(\frac{1}{x+5}=\frac{1}{20}\)

=>\(x+5=20\)

=>\(x=20-5\)

=>\(x=15\)

ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)

=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)

=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)

=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)

=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)

=>\(\frac{1}{x+3}=\frac{1}{20}\)

=>\(x+3=20\)

=>\(x=20-3\)

=>\(x=17\)

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{\left(x+2\right)\left(x+5\right)}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x+2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{3}\left(\frac{1}{2}-\frac{1}{x+5}\right)=\frac{3}{20}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}:\frac{1}{3}\\ \frac{1}{2}-\frac{1}{x+5}=\frac{3}{20}\cdot3\\ \frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{1}{2}\\ \frac{1}{x+5}=\frac{9}{20}+\frac{10}{20}\\ \frac{1}{x+5}=\frac{19}{20}\\ \frac{19}{19\left(x+5\right)}=\frac{19}{20}\\ \Rightarrow19\left(x+5\right)=20\\ 19x+95=20\\ 19x=20-95\\ 19x=-75\\ x=\left(-75\right):19\\ x=\frac{75}{19}\)

Vậy \(x=\frac{75}{19}\)

\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)

Ta có A = 1/2×5 -1/5×8 -1/8×11 -1/11×14 -1/14×17 -1/17*20

=>A3= 3/2×5 -3/5×8 -3/8×11 -3/11×14 -3/14×17 -3/17×20

=>A3= 1/2 -1/5 -1/5 +1/8 -1/8 +1/11 -1/11+1/14 -1/14 +1/17 -1/17 +1/20

=>A3= 1/2 -1/5-1/5+1/20

=>A3= 10/20 -4/20 -4/20 +1/20= 3/20

=>A=3/20:3

=> A =1/20 

Có j ko hiu hỏi mk nha

giúp mk bài dưới dc k

Mk lm câu b trc nha

3 mũ x+1 . 5 mũ y = 3 mũ 2x . 5 mũ x

=> 3 mũ x+1= 3 mũ 2x và 5 mũ y = 5 mũ x

=> x+1=2x và x=y

Từ x+1 =2x => 2x-x=1 => x=1 

Mà x= y=>x=y=1

Vậy.. 

ta có : 1/10 = 1/2.5 

1/40 = 1/5.8

suy ra 1/10 + 1/40 + 1/88 + ... + 1/(3x+2).(3x+5) = 3/20 = 1/2.5 + 1/5.8 + 1/8.11 +....+ 1/(3x+2).(3x+5)

= 1/3(1/2 - 1/5) + 1/3(1/5 - 1/8) + ...+ 1/3(1/8 - 1/11) + ....+ 1/3 [(3x+2) (3x+5)] = 3/20

= 1/3(1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +.... + 1/3x+2 + 1/3x+5) = 3/20

1/3 ( 1/2 - 1/3x+5) = 3/20

suy ra 1/2 - 1/3x+5 = 3/20 : 1/3 = 9/20

suy ra 1/3x+5 = 1/2 - 9/20 = 1/20

suy ra 3x+5 = 20

suy ra 3x = 20 - 5= 15

suy ra x = 15 : 3 = 5

vay x = 5

nhớ k nha!!!!

mk cx nghi nhu tren do

x = 5 đúng rồi đó

ba Nhunh đúng k

bít rùi đó