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.. 

\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)

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

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)

\(\Rightarrow x+5=20\)

\(\Rightarrow x=20-5\)

\(\Rightarrow x=15\)

215

215

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