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Bài 1:
a) \(\sqrt{1-x^2}\)có nghĩa \(\Leftrightarrow\)\(1-x^2\ge0\)
\(\Leftrightarrow\)\(x^2\le1\)
\(\Leftrightarrow\)\(\left|x\right|\le1\)
b) \(\sqrt{\frac{x-2}{x-3}}\)có nghĩa \(\Leftrightarrow\)\(\frac{x-2}{x-3}\ge0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x>3\\x\le2\end{cases}}\)
a:\(M=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)
\(=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
b: \(M=2\sqrt{\sqrt{15+\sqrt{6}}-4}\simeq0.088\)
\(M=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=4\)
\(\Leftrightarrow\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)}^2=4\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|=4\)
Ta có : \(\left|\sqrt{x-4}-2\right|= \left|2-\sqrt{x-4}\right|\)
Áp dụng BĐT \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có :
\(\left|\sqrt{x-4}+2\right|+\left|2-\sqrt{x-4}\right|\ge\left|\sqrt{x-4}+2+2-\sqrt{x-4}\right|=4\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-4}+2\ge0\\2-\sqrt{x-4}\ge0\end{matrix}\right.\Rightarrow x\le8\)
Kết hợp với điều kiện ban đầu \(\Rightarrow4\le x\le8\)
\(a.A=\dfrac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\dfrac{\sqrt{x-2-2.\sqrt{2}.\sqrt{x-2}+2}}{\sqrt{2}}=\dfrac{\sqrt{x-2}-\sqrt{2}}{\sqrt{2}}\) \(b.A=\dfrac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\dfrac{\sqrt{x-2-2.\sqrt{2}.\sqrt{x-2}+2}}{\sqrt{2}}=\dfrac{\sqrt{2}-\sqrt{x-2}}{\sqrt{2}}\)
\(A=\left(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}}\right)\sqrt{x+\sqrt{x^2-32}}\) với \(x\ge4\sqrt{2}\)
Lời giải:
\(A\sqrt{2}=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{2x+\sqrt{(x-4\sqrt{2})(x+4\sqrt{2})}}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})^2}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})\)
\(=(\sqrt{x-4\sqrt{2}})^2-(\sqrt{x+4\sqrt{2}})^2=(x-4\sqrt{2})-(x+4\sqrt{2})=-8\sqrt{2}\)
Lời giải:
\(A\sqrt{2}=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{2x+\sqrt{(x-4\sqrt{2})(x+4\sqrt{2})}}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})^2}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})\)
\(=(\sqrt{x-4\sqrt{2}})^2-(\sqrt{x+4\sqrt{2}})^2=(x-4\sqrt{2})-(x+4\sqrt{2})=-8\sqrt{2}\)
a/ Sai đề.
\(x+2\sqrt{2x-4}=\left(x-2\right)+2.\sqrt{2}.\sqrt{x-2}+2=\left(\sqrt{2}+\sqrt{x-2}\right)^2\)
b/ \(M=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{\left(\sqrt{2}+\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{x-2}\right)^2}\)
\(=\sqrt{2}+\sqrt{x-2}+\left|\sqrt{2}-\sqrt{x-2}\right|\)
1. Nếu \(2\le x\le4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\)
2. Nếu \(x>4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)
ĐKXĐ: x > 4
a, Có \(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
\(\orbr{\begin{cases}=2\sqrt{x-4}\left(với\sqrt{x-4}\ge2\right)\\=4\left(với\sqrt{x-4}< 2\right)\end{cases}}\)
b, Xét \(A=2\sqrt{x-4}\)thì \(\sqrt{x-4}\ge2\)
\(\Leftrightarrow x-4\ge4\)
\(\Leftrightarrow x\ge8\)
Khi đó \(A=2\sqrt{x-4}\ge2\sqrt{8-4}=4\)
Nên \(A_{min}=4\Leftrightarrow x=8\)
c, Với \(x=\sqrt{15+\sqrt{6}}\)thì \(\sqrt{x-4}=\sqrt{\sqrt{15+\sqrt{6}}-4}< 2\)
Nên từ câu a => A = 4