Với 4 < a < 8. Rút gọn \(G=\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}\right)-2}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\frac{4}{a}}\)
\(=\frac{2a}{\sqrt{a-4}}\)
a) \(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
b) \(\sqrt{13-4\sqrt{3}}=\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+1}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}=2\sqrt{3}-1\)
c) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
d) \(\sqrt{3+2\sqrt{2}+\sqrt{6-4\sqrt{2}}}\)
\(=\sqrt{3+2\sqrt{2}+\sqrt{\left(2-\sqrt{2}\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}+2-\sqrt{2}}\)
\(=\sqrt{5+\sqrt{2}}\)
e) \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
\(=2+\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=2+\sqrt{17-4\left(\sqrt{5}+2\right)}\)
\(=2+\sqrt{9-4\sqrt{5}}\)
\(=2+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2+\sqrt{5}-2=\sqrt{5}\)
f) đề sai nhé:
\(\sqrt{3a}.\sqrt{12a}=\sqrt{36a^2}=6a\)\(\left(a\ge0\right)\)
g) \(\sqrt{16a^2b^8}=4b^4\left|a\right|\)
h) \(\sqrt{7a}.\sqrt{63a^3}=\sqrt{441.a^4}=21a^2\)
\(G=\sqrt{a-4+4\sqrt{a-4}+4}+\sqrt{a-4-4\sqrt{a-4}+4}\)
\(=\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}\)
\(=\sqrt{a-4}+2+\sqrt{a-4}-2=2\sqrt{a-4}\)
\(G = \sqrt{a + 4 \sqrt{a – 4}} + \sqrt{a – 4\sqrt{a – 4}} \) \(= \sqrt{a – 4 + 4 + 4\sqrt{a – 4}} + \sqrt{a – 4 + 4 – 4\sqrt{a – 4}}\)
\(= \sqrt{\sqrt{a - 4}^2 + 2^2 + 4\sqrt{a – 4}} + \sqrt{\sqrt{a - 4}^2 + 2^2 - 4\sqrt{a – 4}}\)
\(= \sqrt{(\sqrt{(a – 4)} + 2)^2} + \sqrt{(\sqrt{(a – 4)} - 2)^2}\)
\(= \sqrt{a – 4} + 2 +|\sqrt{a – 4} – 2|\)
+) Với \(4 < a < 8 ⇔ 0 < a – 4 < 4 ⇔ \sqrt{0} < \sqrt{a – 4} < \sqrt{4} ⇔ 0 <\sqrt{a – 4} < 2 \)
Do đó, ta có: \(G = \sqrt{a – 4} + 2 + 2 - \sqrt{a – 4} \) (vì \(2 > \sqrt{a – 4}\))
\(=4\)
➤Với \(4 < a < 8 \) thì \(G = 4 \)