1: \(\dfrac{x}{x-3}+\dfrac{2x}{5\left(x+1\right)}=\dfrac{4x}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow5x\left(x+1\right)+2x\left(x-3\right)=20x\)
\(\Leftrightarrow5x^2+5x+2x^2-6x-20x=0\)
\(\Leftrightarrow7x^2-21x=0\)
\(\Leftrightarrow7x\left(x-3\right)=0\)
=>x=0(nhận) hoặc x=3(loại)
5: \(\Leftrightarrow x^3+8+x^2-2x+4=12\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(x+2\right)\left(x-1\right)=0\)
hay \(x\in\left\{0;1\right\}\)
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