Cho $\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}=1$. Tính $\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}$ 06/09/2021 Bởi Sarah Cho $\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}=1$. Tính $\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}$
Đáp án: Ta có : $(\sqrt[]{x^2-6x+13} – \sqrt[]{x^2-6x+10}) . (\sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10})$ $ = (\sqrt[]{x^2-6x+13})^2 – (\sqrt[]{x^2-6x+10})^2$ ` = x^2 – 6x + 13 – (x^2 – 6x +10)` ` = x^2 – 6x + 13 – x^2 + 6x – 10` ` = 3` $ (\sqrt[]{x^2-6x+13} – \sqrt[]{x^2-6x+10}) . (\sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10}) = 3$ `=>`$ 1. (\sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10}) = 3$ ( do $\sqrt[]{x^2-6x+13} – \sqrt[]{x^2-6x+10} = 1$ ) ` =>`$ \sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10} = 3$ Bình luận
Đáp án:
Ta có :
$(\sqrt[]{x^2-6x+13} – \sqrt[]{x^2-6x+10}) . (\sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10})$
$ = (\sqrt[]{x^2-6x+13})^2 – (\sqrt[]{x^2-6x+10})^2$
` = x^2 – 6x + 13 – (x^2 – 6x +10)`
` = x^2 – 6x + 13 – x^2 + 6x – 10`
` = 3`
$ (\sqrt[]{x^2-6x+13} – \sqrt[]{x^2-6x+10}) . (\sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10}) = 3$
`=>`$ 1. (\sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10}) = 3$
( do $\sqrt[]{x^2-6x+13} – \sqrt[]{x^2-6x+10} = 1$ )
` =>`$ \sqrt[]{x^2-6x+13} + \sqrt[]{x^2-6x+10} = 3$
Đáp án:
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