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\(VT=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)
\(=\sqrt{\left(\sqrt{a-2}\right)^2+4\sqrt{a-2+4}}+\sqrt{\left(\sqrt{a}-2\right)^2-4\sqrt{a-2}+4}\)
\(=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
\(=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)
Nếu \(a=6\) thì \(VT=\sqrt{6-2}+2+\sqrt{6-2}-2=4\)
Nếu \(2\le a< 6\) thì \(VT=\sqrt{a-2}+2+2-\sqrt{a-2}=4\)
\(\left(\frac{a-4}{a}\right)\left(\frac{\sqrt{a}-1}{\sqrt{a}+2}-\frac{\sqrt{a}+1}{\sqrt{a}-2}\right)=\left(\frac{a-4}{a}\right)\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\)
\(=\left(\frac{a-4}{a}\right)\left(\frac{a-3\sqrt{a}+2-a-3\sqrt{a}-2}{a-4}\right)\)
\(=\frac{-6\sqrt{a}}{a}=\frac{-6}{\sqrt{a}}\)
a,\(\sqrt{23-8\sqrt{7}}-\sqrt{7}=\sqrt{16-8\sqrt{7}+7}-\sqrt{7}=\sqrt{\left(4-\sqrt{7}\right)^2}-\sqrt{7}=\left|4-\sqrt{7}\right|-\sqrt{7}=4-\sqrt{7}-\sqrt{7}=4\)
\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=4\)
\(\Leftrightarrow\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}=4\)
\(\Leftrightarrow\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|=4\)
Ta thấy :
\(VT=\left|\sqrt{a-2}+2\right|+\left|2-\sqrt{a-2}\right|\ge\left|\sqrt{a-2}+2+2-\sqrt{a-2}\right|=4\)
\(\Rightarrow VT\ge4\)
Dấu "=" xảy ra khi \(\left(\sqrt{a-2}+2\right)\left(2-\sqrt{a-2}\right)\ge0\Rightarrow a\le4\)
Kém theo ĐKXĐ ta tìm đc \(2\le a\le4\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
Phải có ĐK là \(a\le2\le6\) bạn nhé
Ta có
\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)
\(=\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)
\(=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)
\(=\sqrt{a-2}+2+\left|\sqrt{a-2}-2\right|\)
\(=\sqrt{a-2}+2+2-\sqrt{a-2}=4\)
gõ nhầm ĐK là \(2\le a\le6\)