a,
\(-5x+3x^2=0\\ \Leftrightarrow3x^2-5x=0\\ \Leftrightarrow x\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0;\frac{5}{3}\right\}\)
b,\(5\left(x^2-2x\right)=\left(3+5x\right)\left(x-1\right)\)
\(\Leftrightarrow5x^2-10x=3x+5x^2-3-5x\)
\(\Leftrightarrow5x^2-10x-3x-5x^2+5x=-3\)
\(\Leftrightarrow-8x=-3\Leftrightarrow x=\frac{3}{8}\)
Vậy phương trình có 1 nghiệm duy nhất là \(x=\frac{3}{8}\)
c, \(\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow16x^2+24x+9=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32+5=0\)
\(\Leftrightarrow\left(12x^2+2x\right)+\left(30x+5\right)=0\)
\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{1}{6};-\frac{5}{2}\right\}\)
a/\(\Leftrightarrow x\left(3x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)
b/\(\Leftrightarrow5x^2-10x=3x-3+5x^2-5x\Leftrightarrow8x-3=0\Rightarrow x=\frac{3}{8}\)
c/ \(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\Leftrightarrow12x^2+32x+5=0\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\Leftrightarrow\left(2x+5\right)\left(6x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=-\frac{5}{2}\\x=-\frac{1}{6}\end{matrix}\right.\)
