Question

câu 1 giải các pt sau

a,3x-12=0        b,(x-2)(2x+3)=0           c,\(\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\)

Answer 1

\(a,3x-12=0\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=4\)

\(b,\left(x-2\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\left(dkxd:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2-6\left(x-2\right)-x^2}{x^2-4}=0\)

\(\Leftrightarrow x^2+4x+4-6x+12-x^2=0\)

\(\Leftrightarrow-2x+16=0\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\left(tmdk\right)\)

Answer 2

\(a,3x-12=0\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=4.\)

Vậy \(S=\left\{4\right\}\)

\(b,\left(x-2\right)\left(2x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-3}{2}.\end{matrix}\right.\)

Vậy \(S=\left\{2;\dfrac{-3}{2}\right\}\)

\(c,\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\left(ĐKXĐ:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Rightarrow x^2+4x+4-6x+12-x^2=0\)

\(\Leftrightarrow-2x+16=0\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\left(tm\right).\)

Vậy \(S=\left\{8\right\}\)