a
ĐK: \(x\ne5\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\\ \Leftrightarrow\dfrac{x-5}{3}=\dfrac{12}{x-5}\\ \Leftrightarrow\left(x-5\right)^2=12.3=36\\ \Leftrightarrow\left\{{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=11\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
b
ĐK: \(x\ne0;x\ne-1\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x}.\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2023}{4048}\\ \Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2023}{4048}=\dfrac{1}{4048}\\ \Leftrightarrow4048=x+1\\ \Leftrightarrow x=4047\left(tm\right)\)
a: =>(x-5)/3=12/(x-5)
=>(x-5)^2=36
=>x-5=6 hoặc x-5=-6
=>x=11 hoặc x=-1
b: =>\(2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2023}{2024}\)
=>1/2-1/3+1/3-1/4+...+1/x-1/x+1=2023/4048
=>1/2-1/x+1=2023/4048
=>1/(x+1)=1/4048
=>x+1=4048
=>x=4047
