a: \(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{-4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow x^2-8x+12=-32\)
\(\Leftrightarrow x^2-8x+16+28=0\)
\(\Leftrightarrow\left(x-4\right)^2+28=0\)(vô lý)
c: \(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)
=>x-59=0
hay x=59
a, ĐKXĐ:\(x\ne2,x\ne3,x\ne4,x\ne5,x\ne6\)
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}-\dfrac{1}{8}=0\\ \Leftrightarrow\dfrac{8\left(x-6\right)-8\left(x-2\right)-\left(x-2\right)\left(x-6\right)}{8\left(x-2\right)\left(x-6\right)}=0\)
\(\Rightarrow8x-48-8x+16-x^2+8x-12=0\\ \Leftrightarrow-x^2+8x-44=0\\ \Leftrightarrow x^2-8x+44=0\\\left(x^2-8x+16\right)+28=0\\ \Leftrightarrow\left(x-4\right)^2+28=0\left(vô.lí\right)\)
a, ĐKXĐ:\(x\ne2,x\ne3,x\ne4,x\ne5,x\ne6\)
\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}-\dfrac{1}{8}=0\\ \Leftrightarrow\dfrac{8\left(x-6\right)-8\left(x-2\right)-\left(x-2\right)\left(x-6\right)}{8\left(x-2\right)\left(x-6\right)}=0\)
\(\Rightarrow8x-48-8x+16-x^2+8x-12=0\\ \Leftrightarrow-x^2+8x-44=0\\ \Leftrightarrow x^2-8x+44=0\\\left(x^2-8x+16\right)+28=0\\ \Leftrightarrow\left(x-4\right)^2+28=0\left(vô.lí\right)\)
\(b,\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\\ \Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\\ \Leftrightarrow\dfrac{x-259}{17}+\dfrac{x-259}{19}+\dfrac{x-259}{21}+\dfrac{x-259}{23}=0\\ \Leftrightarrow\left(x-259\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\\ \Leftrightarrow x=259\left(vì.\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\ne0\right)\)\(c,\dfrac{x-29}{30}+\dfrac{x-30}{29}=\dfrac{29}{x-30}+\dfrac{30}{x-29}\\ \Leftrightarrow\left(\dfrac{x-29}{30}-1\right)+\left(\dfrac{x-30}{29}-1\right)=\left(\dfrac{29}{x-30}-1\right)+\left(\dfrac{30}{x-29}-1\right)\\ \Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{59-x}{x-30}+\dfrac{59-x}{x-29}\\ \Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=-\dfrac{x-59}{x-30}-\dfrac{x-59}{x-29}\\ \Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}+\dfrac{x-59}{x-30}+\dfrac{x-59}{x-29}=0\)
\(\Leftrightarrow\left(x-59\right)\left(\dfrac{1}{30}+\dfrac{1}{29}+\dfrac{1}{x-30}+\dfrac{1}{x-29}\right)=0\\ \Leftrightarrow x=59\left(vì.\dfrac{1}{30}+\dfrac{1}{29}+\dfrac{1}{x-30}+\dfrac{1}{x-29}\ne0\right)\)
