giải các phương trình:
\(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
\(\dfrac{x+2}{2020}+\dfrac{x+4}{2018}=\dfrac{x+6}{2016}+\dfrac{x+8}{2014}\)
Giúp tớ với, thầy réo tớ kinh lắm rồi!
\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)x-2-5(x+1)=15
\(\Leftrightarrow\) x-2-5x-5=15
\(\Leftrightarrow\)x-5x=15+2+5
\(\Leftrightarrow\)-4x=22
\(\Leftrightarrow\)x=-\(\dfrac{11}{2}\)
vậy
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