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b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)
\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)
\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)
\(\Leftrightarrow2x^2+9x-11=0\)
\(\Leftrightarrow2x^2+11x-2x-11=0\)
\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)
\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)
b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)
\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)
\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)
\(\Leftrightarrow12x^2+19x-18=0\)
\(\Leftrightarrow12x^2+27x-8x-18=0\)
\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)
\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)
