Question

câu 1. đơn giản biểu thức

a. A = \(\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}\)

b. B = \(\left(\sqrt{3-\sqrt{7}}\right)^6\)

c. C = \(\left(\sqrt{\sqrt{2}-\sqrt{3}}\right)^2\)

d. D = \(\sqrt{3+2\sqrt{2}}\)

e. E = \(\sqrt{8-2\sqrt{15}}\)

Answer 1

d, \(D=\sqrt{3+2\sqrt{2}}=\sqrt{2+2.\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

e,\(E=\sqrt{8-2\sqrt{15}}=\sqrt{5-2.\sqrt{5}.\sqrt{3}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}\)

Answer 2

a,ĐKXĐ: \(\forall x\in R\)

\(\Rightarrow A=\left|a+3\right|+\left|a-3\right|\)\(=\left|-a-3\right|+\left|a-3\right|\)

Vì \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) *Dấu ''='' xảy ra\(\Leftrightarrow A.B\ge0\) *

\(\Rightarrow A\ge\left|-a-3+a-3\right|=6\)

Dấu ''='' xảy ra \(\Leftrightarrow\left(-a-3\right)\left(a-3\right)\ge0\Leftrightarrow\left(a+3\right)\left(a-3\right)\ge0\)

\(\Leftrightarrow-3\le a\le3\)

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