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電子課本網(wǎng) 第4頁(yè)

第4頁(yè)

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解:對(duì)于方程$x^{2}-\frac{1}{2}=0,$移項(xiàng)可得$x^{2}=\frac{1}{2}。$直接開(kāi)平方,$x = \pm\sqrt{\frac{1}{2}}=\pm\frac{\sqrt{2}}{2}。$所以$x_{1}=\frac{\sqrt{2}}{2},$$x_{2}=-\frac{\sqrt{2}}{2}。$
解:對(duì)于方程$4y^{2}=0,$兩邊同時(shí)除以$4,$得$y^{2}=0。$直接開(kāi)平方,$y = 0。$所以$y_{1}=y_{2}=0。$
解:對(duì)于方程$2x^{2}-98 = 0,$移項(xiàng)得$2x^{2}=98,$兩邊同時(shí)除以$2,$$x^{2}=49。$直接開(kāi)平方,$x=\pm\sqrt{49}=\pm7。$所以$x_{1}=7,$$x_{2}=-7。$
解:對(duì)于方程$(x + 3)^{2}=2,$直接開(kāi)平方,$x + 3=\pm\sqrt{2}。$移項(xiàng)得$x=-3\pm\sqrt{2}。$所以$x_{1}=-3+\sqrt{2},$$x_{2}=-3-\sqrt{2}。$
解:對(duì)于方程$(x + 2)^{2}=(2x + 3)^{2},$直接開(kāi)平方,$x + 2=\pm(2x + 3)。$當(dāng)$x + 2=2x + 3$時(shí),移項(xiàng)得$x-2x=3 - 2,$即$-x = 1,$解得$x=-1。$當(dāng)$x + 2=-(2x + 3)$時(shí),去括號(hào)得$x + 2=-2x-3,$移項(xiàng)得$x + 2x=-3 - 2,$$3x=-5,$解得$x=-\frac{5}{3}。$所以$x_{1}=-1,$$x_{2}=-\frac{5}{3}。$
解:對(duì)于方程$100(1 - x)^{2}=64,$兩邊同時(shí)除以$100,$得$(1 - x)^{2}=\frac{64}{100}。$直接開(kāi)平方,$1 - x=\pm\frac{8}{10}=\pm\frac{4}{5}。$當(dāng)$1 - x=\frac{4}{5}$時(shí),移項(xiàng)得$x=1-\frac{4}{5}=\frac{1}{5}。$當(dāng)$1 - x=-\frac{4}{5}$時(shí),移項(xiàng)得$x=1+\frac{4}{5}=\frac{9}{5}。$所以$x_{1}=\frac{1}{5},$$x_{2}=\frac{9}{5}。$
$x_1=13,$$x_2=-13$
$x_1=x_2=-3$