Question

Answer 1

\(a,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\\ \Leftrightarrow3x^2-7x+15x-35=15-2x^2+3x-10x\\ \Leftrightarrow3x^2+2x^2-7x+15x-3x+10x-35-15=0\\ \Leftrightarrow5x^2+15x-50=0\\ \Leftrightarrow x^2+3x-10=0\\ \Leftrightarrow x^2+5x-2x-10=0\\ \Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\\ Vậy:S=\left\{-5;2\right\}\\ b,\dfrac{-x+3}{2}=\dfrac{x-2}{3}\\ \Rightarrow3\left(3-x\right)=2\left(x-2\right)\\ \Leftrightarrow9-3x=2x-4\\ \Leftrightarrow2x+3x=9+4\\ \Leftrightarrow5x=13\\ \Leftrightarrow x=\dfrac{13}{5}\\ Vậy:S=\left\{\dfrac{13}{5}\right\}\)

Answer 2

\(c,\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\left(ĐKXĐ:x\ne\pm2\right)\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(x-1\right)+5\left(x-2\right)}{x^2-4}=\dfrac{12+\left(x^2-4\right)}{x^2-4}\\ \Leftrightarrow x^2+2x-x-2+5x-10=12+x^2-4\\ \Leftrightarrow x^2-x^2+2x-x+5x=2+10+12-4\\ \Leftrightarrow6x=20\\ \Leftrightarrow x=\dfrac{10}{3}\left(TM\right)\\ Vậy:S=\left\{\dfrac{10}{3}\right\}\)

Answer 3

<=>3x^2+15x-7x-35=15-10x+3x-2x^2

<=>3x^2+2x^2-7x-3x+10x+15x=15+35

<=>5x^2+15x-50=0

<=>5(x^2+3x-10)=0

<=>x^2-2x+5x-10=0

<=>X=-5,x=2

Answer 4

a) \(\text{( 3x - 7 ) ( x + 5 ) = ( 5 + x ) ( 3 - 2x )}\)

\(\Leftrightarrow\) \(\text{( 3x - 7 ) (x + 5 ) - ( 5 + x ) ( 3 - 2x ) = 0}\)

\(\Leftrightarrow\) \(\text{( x + 5 ) [( 3x - 7 0 - ( 3 - 2x )] = 0}\)

\(\Leftrightarrow\) \(\text{( x + 5 ) ( 3x - 7 - 3 + 2x ) = 0}\)

\(\Leftrightarrow\) \(\text{( x + 5 ) ( 5x - 10 ) = 0}\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\text{x + 5 = 0}\\\text{5x - 10 = 0}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\text{x = -5 }\\\text{x = 2}\end{matrix}\right.\)

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b) \(\dfrac{\text{-x + 3}}{\text{2}}=\dfrac{\text{x - 2 }}{\text{3}}\)

\(\Leftrightarrow\) \(\text{-3x + 9 = 2x - 4}\)

\(\Leftrightarrow\) \(\text{-5x = -13}\)

\(\Leftrightarrow\) \(\text{x =}\dfrac{\text{13}}{\text{5}}\)

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c) \(\dfrac{\text{x - 1}}{\text{x - 2}}+\dfrac{\text{5}}{\text{x + 2}}=\dfrac{\text{12}}{\text{x^2 - 4}}+1\) ( ĐK : \(\text{x}\) \(\ne\pm\) \(\text{2}\) )

\(\Leftrightarrow\) \(\text{6x - 20 = 0}\)

\(\Leftrightarrow\) \(\text{x = }\dfrac{\text{10}}{\text{3}}\)