\(\dfrac{2x+1}{6}=\dfrac{3-x}{9}\)

\(\Leftrightarrow9\left(2x+1\right)=6\left(3-x\right)\)

\(\Leftrightarrow18x+9=18-6x\)

\(\Leftrightarrow24x=9\)

\(\Leftrightarrow x=\dfrac{3}{8}\)

Bn tham khảo đây nhé: https://diendantoanhoc.org/topic/140204-t%C3%A0i-li%E1%BB%87u-d%C3%A3y-s%E1%BB%91/

1) Có \(u_{n+1}-u_n=\dfrac{1}{2}u^2_n-2u_n+2=\dfrac{1}{2}\left(u_n-2\right)^2\) (1)

+) CM \(u_n>2\) (n thuộc N*)

n=1 : u₁= 5/2 > 2 (đúng)

Giả sử n=k, u_k > 2 (k thuộc N*)

Ta cần CM n = k + 1. Thật vậy ta có:

\(u_{k+1}=\dfrac{1}{2}u^2_k-u_k+2=\dfrac{1}{2}\left(u_k-2\right)^2+u_k\) (đúng)

Vậy u_n > 2 (n thuộc N*)        (2)

Từ (1) (2) => u_n+1 - u_n > 0, hay u_n+1 > u_n

=> (u_n) là dãy tăng => \(\lim\limits_{n\rightarrow\infty}u_n=+\infty\)

2) \(2u_{n+1}=u^2_n-2u_n+4\)

\(\Leftrightarrow2u_{n+1}-4=u^2_n-2u_n\)

\(\Leftrightarrow2\left(u_{n+1}-2\right)=u_n\left(u_n-2\right)\)

\(\Leftrightarrow\dfrac{1}{u_{n+1}-2}=\dfrac{2}{u_n\left(u_n-2\right)}=\dfrac{1}{u_n-2}-\dfrac{1}{u_n}\)

\(\Leftrightarrow\dfrac{1}{u_n}=\dfrac{1}{u_n-2}-\dfrac{1}{u_{n+1}-2}\)

\(S=\dfrac{1}{u_1}+\dfrac{1}{u_2}+...+\dfrac{1}{u_n}\)

\(=\dfrac{1}{u_1-2}-\dfrac{1}{u_2-2}+\dfrac{1}{u_2-2}+...-\dfrac{1}{u_{n+1}-2}\)

\(=\dfrac{1}{u_1-2}-\dfrac{1}{u_{n+1}-2}\)

\(=2-\dfrac{1}{u_{n+1}-2}\)

\(\Leftrightarrow\lim\limits_{n\rightarrow\infty}S=2\)