找出下列式子中的错误并改正。
(a) \( (2 x+5)^{2}=4 x^{2}+25 \)
(b) \( \left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=x^{2}-\frac{1}{4} \)
(c) \( (5 a-b)^{2}=10 a^{2}-5 a b+b^{2} \)
(d) \( (p-3)(p-7)=p^{2}+21 \)
待办事项
我们需要更正以下表达式的错误。
(a) \( (2 x+5)^{2}=4 x^{2}+25 \)
(b) \( \left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=x^{2}-\frac{1}{4} \)
(c) \( (5 a-b)^{2}=10 a^{2}-5 a b+b^{2} \)
(d) \( (p-3)(p-7)=p^{2}+21 \)
解答
我们知道,
$(a+b)^2=a^2+2ab+b^2$
$(a-b)^2=a^2-2ab+b^2$
因此,
(a) 左侧 $=(2x+5)^2$
$=(2x)^2+2.(2x).(5)+(5)^2$
$=4x^2+20x+25$
右侧 $=4x^2+25$
因此,我们需要在右侧加上 $20x$
(b) 左侧 $=(x-\frac{1}{2})(x-\frac{1}{2})$
$=(x-\frac{1}{x})^2$
$=x^2-2.x.\frac{1}{x}+(\frac{1}{x})^2$
$=x^2-2+\frac{1}{x^2}$
右侧 $=x^{2}-\frac{1}{4}$
因此,我们需要将右侧的 $-\frac{1}{4}$ 替换为 $-2+\frac{1}{x^2}$
(c) 左侧 $=(5 a-b)^{2}$
$=(5a)^2-2.5a.b+(b)^2$
$=25a^2-10ab+b^2$
右侧 $=10 a^{2}-5 a b+b^{2}$
因此,我们需要将右侧的 $10 a^{2}-5 a b$ 替换为 $25a^2-10ab$。
(d) 左侧 $=(p-3)(p-7)$
$=p(p)-p(7)-3(p)-3(-7)$
$=p^2-7p-3p+21$
$=p^2-10p+21$
右侧 $=p^{2}+21$
因此,我们需要在右侧加上 $-10p$。
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