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信息发布者:
D
C
$x_1 = \sqrt{6},x_2 = -\sqrt{6}$
$x_1 = 2,x_2 = -4$
解:
$\begin{aligned}\frac{1}{13}m^{2}&=0\\m^{2}&=0\\m_1&=m_2 = 0\end{aligned}$
解:
$\begin{aligned}9x^{2}-0.16&=0\\9x^{2}&=0.16\\x^{2}&=\frac{0.16}{9}\\x&=\pm\sqrt{\frac{0.16}{9}}\\x_1&=\frac{2}{15},x_2 = -\frac{2}{15}\end{aligned}$
解:
$\begin{aligned}3(x + 4)^{2}&=15\\(x + 4)^{2}&=5\\x+4&=\pm\sqrt{5}\\x_1&=\sqrt{5}-4,x_2 = -\sqrt{5}-4\end{aligned}$
解:
$\begin{aligned}(2x + 3)^{2}&=(3x + 2)^{2}\\2x+3&=\pm(3x + 2)\\\end{aligned}$
当$2x + 3 = 3x + 2$时,$3x-2x=3 - 2,$
解得$x_1 = 1;$
当$2x + 3 = -(3x + 2)$时,$2x+3=-3x - 2,$
$2x + 3x=-2 - 3,$
$5x=-5,$
解得$x_2=-1。$
7(答案不唯一)
解:当$m = 7$时,方程为$(x - 2)^{2}=16 - 7,$即$(x - 2)^{2}=9。$
$\begin{aligned}x-2&=\pm3\\x_1&=5,x_2=-1\end{aligned}$
A
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