điều kiện : \(x\ne5\)
ta có : \(\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{1}{x-5}=4\)
\(\Leftrightarrow\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{x-n}{x-5}=4\) (với \(n=x-1\))
\(\Leftrightarrow\dfrac{x-5+x-6+x-7+...+x-n}{x-5}=4\)
\(\Leftrightarrow\dfrac{\left(n-4\right)x-5\left(n-4\right)-\left(0+1+2+3+...+n-5\right)}{x-5}=4\)
\(\Leftrightarrow\dfrac{\left(x-5\right)x-5\left(x-5\right)-\left(0+1+2+3+...+x-6\right)}{x-5}=4\)
\(\Leftrightarrow\dfrac{\left(x-5\right)x-5\left(x-5\right)-\dfrac{\left(x-6\right)\left(x-5\right)}{2}}{x-5}=4\)
\(\Leftrightarrow x-5-\dfrac{x-6}{2}=4\Leftrightarrow\dfrac{2x-10-x+6}{2}=4\)
\(\Leftrightarrow\dfrac{x-4}{2}=4\Leftrightarrow x-4=2\Leftrightarrow x=6\left(tmđk\right)\)
vậy \(x=6\)
ĐK: \(x\ge5\)
\(\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{1}{x-5}=4\)
\(\Rightarrow\dfrac{x-5}{x-5}+\dfrac{x-6}{x-5}+\dfrac{x-7}{x-5}+...+\dfrac{x-\left(x-1\right)}{x-5}=4\)
\(\Rightarrow\dfrac{\left(x-5\right)+\left(x-6\right)+\left(x-7\right)+...+\left[x-\left(x-1\right)\right]}{x-5}=4\) (1)
Số số hạng của tử số của vế trái của (1) là: \(x-1-5+1=x-5\)
\(\Rightarrow\dfrac{x\left(x-5\right)-\left(x-1+5\right)\left(x-5\right):2}{x-5}=4\)
\(\Rightarrow x-\left(x+4\right):2=4\)
\(\Rightarrow x-\dfrac{1}{2}x-2=4\)
\(\Rightarrow\dfrac{1}{2}x=6\)
\(\Rightarrow x=12\)