Đáp án + Giải thích các bước giải:
`1, (10 + 7x).(x - 1) = (9x - 2).(x - 1)`
`<=> (10 + 7x).(x - 1) - (9x - 2).(x - 1) =0`
`<=> (x-1)(10+7x-9x+2)=0`
`<=> (x-1)(12-2x)=0`
`<=> (x-1).2(6-x)=0`
`<=>` \(\left[ \begin{array}{l}x-1=0\\6-x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l} x=1\\x=6\end{array} \right.\)
`2, (9x - 4).(x - 1/2) - (x - 1/2).(6 + x) = 0`
`<=> (x - 1/2)(9x-4-6-x)=0`
`<=> (x - 1/2)(8x-10)=0`
`<=>` \(\left[ \begin{array}{l} x- \dfrac{1}{2} =0\\8x-10=0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l} x=\dfrac{1}{2} \\x=\dfrac{5}{4} \end{array} \right.\)
`3, 9x^2 - 1 = (3x - 1).(x + 4)`
`<=> (3x-1).(3x+1)=(3x - 1).(x + 4)`
`<=> (3x-1).(3x+1)-(3x - 1).(x + 4)=0`
`<=> (3x-1).(3x+1-x-4)=0`
`<=> (3x-1).(2x-3)=0`
`<=>` \(\left[ \begin{array}{l} 3x-1=0\\2x-3=0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l} x=\dfrac{1}{3} \\x=\dfrac{3}{2} \end{array} \right.\)
`4, (x + 7).(3x + 1) = 49 - x^2`
`<=> (x + 7).(3x + 1) =(7-x).(7+x)`
`<=> (x + 7).(3x + 1) - (7-x).(7+x)=0`
`<=> (x+7).(3x+1-7+x)=0`
`<=> (x+7).(4x-6)=0`
`<=>` \(\left[ \begin{array}{l}x+7=0\\4x-6=0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-7\\x=\dfrac{3}{2} \end{array} \right.\)
`5, (2x + 1)^2 = (x - 1)^2`
`<=> (2x + 1)^2 - (x - 1)^2 =0`
`<=> (2x+1-x+1)(2x+1+x-1)=0`
`<=> (x+2).3x=0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\3x=0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\)
`6, x^3 - 5x^2 + 6x = 0`
`<=> x(x^2-5x+6)=0`
`<=> x.[x(x-2)-3(x-2)]=0`
`<=> x(x-2)(x-3)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x-2=0\\x-3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\x=2\\ x=3\end{array} \right.\)
`7, 3x^2 + 5x + 2 = 0`
`<=> 3x(x+1)+2(x+1)=0`
`<=> (3x+2)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}3x+2=0\\x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l} x=\dfrac{-2}{3} \\x=-1\end{array} \right.\)
