Đáp án `+` Giải thích các bước giải `!`
`to` Tìm `x:`
`a)`
`(x+3)^2+(x-2)^2 = 2x^2`
`<=> x^2+6x+9+x^2-4x+4-2x^2 = 0`
`<=> (x^2+x^2-2x^2)+(6x-4x)+(9+4) = 0`
`<=> 2x+13 = 0`
`<=> 2x= -13`
`<=>x=-(13)/2`
Vậy `S= {-(13)/2}`
`b)`
`4x^2-x = 0`
`<=> x(4x-1) = 0`
`<=>` \(\left[ \begin{array}{l}x=0\\4x-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\4x=1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{4}\end{array} \right.\)
Vậy `S= {0; 1/4}`
`c)`
`7x(x-2) = x-2`
`<=> 7x(x-2)-x+2 = 0`
`<=> 7x(x-2)-(x-2) = 0`
`<=> (7x-1)(x-2) = 0`
`<=>` \(\left[ \begin{array}{l}7x-1=0\\x-2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}7x=1\\x=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{7}\\x=2\end{array} \right.\)
Vậy `S= {1/7; 2}`
`d)`
`(3x-5)^2-x(3x-5) = 0`
`<=> (3x-5)(3x-5-x) = 0`
`<=> (3x-5)(2x-5) = 0`
`<=>` \(\left[ \begin{array}{l}3x-5=0\\2x-5=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=5\\2x=5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=\dfrac{5}{2}\end{array} \right.\)
Vậy `S= {5/3; 5/2}`
