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a.\(\dfrac{x}{3}+\dfrac{2x-1}{5}=2-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{5x+3\left(2x-1\right)}{15}=\dfrac{30-5x}{15}\)
\(\Leftrightarrow5x+3\left(2x-1\right)=30-5x\)
\(\Leftrightarrow5x+6x-3-30+5x=0\)
\(\Leftrightarrow16x-33=0\)
\(\Leftrightarrow16x=33\)
\(\Leftrightarrow x=\dfrac{33}{16}\)
b.\(\left(2x-1\right)\left(3x+2\right)=\left(5x-8\right)\left(2x-1\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)-\left(5x-8\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(10-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=5\end{matrix}\right.\)
c.\(ĐK:x\ne-1;3\)
\(\Rightarrow\dfrac{x-1}{x+1}+\dfrac{x+5}{x-3}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)+\left(x+5\right)\left(x+1\right)}{\left(x+1\right)\left(x-3\right)}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)+\left(x+5\right)\left(x+1\right)=8\)
\(\Leftrightarrow x^2-3x-x+3+x^2+x+5x+5-8=0\)
\(\Leftrightarrow2x^2+2x=0\)
\(\Leftrightarrow2x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
a: \(\Leftrightarrow5x+3\left(2x-1\right)=30-5x\)
=>5x+6x-3=30-5x
=>11x-3=30-5x
=>16x=33
hay x=33/16
b: \(\Leftrightarrow\left(2x-1\right)\left(5x-8\right)-\left(2x-1\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(5x-8-3x-2\right)=0\)
=>(2x-1)(2x-10)=0
=>x=1/2 hoặc x=5
c: \(\Leftrightarrow\dfrac{x-1}{x+1}+\dfrac{x+5}{x-3}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x^2-4x+3+x^2+6x+5=8\)
\(\Leftrightarrow2x^2+2x=0\)
=>2x(x+1)=0
=>x=0(nhận) hoặc x=-1(loại)