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第8页

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解：$x(x - 4)=0$

$x = 0$或$x - 4 = 0$

$x_{1} = 0，$$x_{2} = 4$

解：$(2y + 1)(y - 2)=0$

$2y + 1 = 0$或$y - 2 = 0$

$y_{1} = -\frac {1}{2}，$$y_{2} = 2$

解：$2(x - 4)+x(x - 4)=0$

$(x - 4)(2 + x)=0$

$ x - 4 = 0$或$2 + x = 0$

$x_{1} = 4，$$x_{2} = - 2$

解：$4x(5x - 2)=3(5x - 2)$

$(5x - 2)(4x - 3)=0$

$5x - 2 = 0$或$4x - 3 = 0$

$x_{1}=\frac {2}{5}，$$x_{2}=\frac {3}{4}$

 解：由​$x^2 - 49 = 0$​得​$x^2=49，$​

所以​$x=\pm\sqrt{49}=\pm7，$

​即​$x_1 = 7，$​​$x_2 = - 7。$​

 解：由​$(x + 1)^2 - 25 = 0$​得​$(x + 1)^2=25，$​

 则​$x + 1=\pm\sqrt{25}=\pm5。$​

 当​$x + 1 = 5$​时，解得​$x_1 = 4；$​

 当​$x + 1 = - 5$​时，解得​$x_2 = - 6。$​

解：$\frac {1}{2}(x - 2)^2 - 8 = 0$

$(x - 2)^2 = 16$

$ x - 2=\pm \sqrt {16}=\pm 4$

$ $当$x - 2 = 4$时，解得$x_{1} = 6$

$ $当$x - 2 = - 4$时，解得$x_{2} = - 2$

解：$(2x - 1)^2 - x^2 = 0$

$(2x - 1 + x)(2x - 1 - x)=0$

$(3x - 1)(x - 1)=0$

$x_{1}=\frac {1}{3}，$$x_{2} = 1$