\(a,6x-4=5x\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\\ b,\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\\ \Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\\ \Leftrightarrow4x+6=15-12x\\ \Leftrightarrow16x-9=0\\ \Leftrightarrow x=\dfrac{9}{16}\\ c,\left(x+7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)
d, ĐKXĐ:\(x\ne\pm3\)
\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x+5}{\left(x+3\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{2x+6+3x-9-3x-5}{\left(x+3\right)\left(x-3\right)}=0\\ \Rightarrow2x-8=0\\ \Leftrightarrow x=4\left(tm\right)\)
a.6x-4=5x <=> x=4
b.\(\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\)
\(\Leftrightarrow\dfrac{2\left(2x+3\right)}{6}=\dfrac{3\left(5-4x\right)}{6}\)
\(\Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\)
\(\Leftrightarrow4x+6=15-12x\)
\(\Leftrightarrow16x=11\)
\(\Leftrightarrow x=\dfrac{11}{16}\)
c.(x+7)(x-10)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)
d.\(ĐK:x\ne\pm3\)
\(\Rightarrow\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{2\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow2\left(x+3\right)+3\left(x-3\right)=3x+5\)
\(\Leftrightarrow2x+6+3x-9-3x-5=0\)
\(\Leftrightarrow2x-8=0\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\left(tm\right)\)
a, \(6x-5x=4\Leftrightarrow x=4\)
b, \(4x+6=15-12x\Leftrightarrow16x=9\Leftrightarrow x=\dfrac{9}{16}\)
c, \(\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)
d, đk : x khác -3 ; 3
\(2x+6+3x-9=3x+5\Leftrightarrow2x=8\Leftrightarrow x=4\left(tmđk\right)\)
\(a.6x-4=5x\)
\(\Leftrightarrow x=4\)
\(b.\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\)
\(\Leftrightarrow\dfrac{4x+6}{6}=\dfrac{15-12x}{6}\)
\(\Leftrightarrow4x+6=15-12x\)
\(\Leftrightarrow16x=9\)
\(\Leftrightarrow x=\dfrac{9}{16}\)
\(c.\left(x+7\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)
\(d.\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\left(đk:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x+6}{x^2-9}+\dfrac{3x-9}{x^2-9}=\dfrac{3x+5}{x^2-9}\)
\(\Rightarrow2x+6+3x-9=3x+5\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\left(tm\right)\)
