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- a.\(A=\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(\sqrt{2}A=\sqrt{12+8\sqrt{2}}+\sqrt{12-8\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{2}+2\right)^2}+\sqrt{\left(2\sqrt{2}-2\right)^2}\)
\(=2\sqrt{2}+2+2\sqrt{2}-2=4\sqrt{2}\)
\(A=\frac{4\sqrt{2}}{\sqrt{2}}=4\)
Bài 1:
a) \(\sqrt{6+4\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}+\left|2-\sqrt{2}\right|\)
\(=2+\sqrt{2}+2-\sqrt{2}\)( Vì \(2>\sqrt{2}\))
\(=4\)
b) Hình như sai đầu bài
Bài 2
Ta có \(VP=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2=VT\)
a)\(B=\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{1}{\sqrt{0}+\sqrt{4}}=\frac{1}{2}\)
b)\(M=A+B=\frac{2\sqrt{y}}{x-y}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)\(=\frac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{1}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{2\sqrt{y}+\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)\(=\frac{2\sqrt{y}+2\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(=\frac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(=\frac{2}{\sqrt{x}-\sqrt{y}}\)
c)\(M=\frac{2}{\sqrt{x}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{4y}-\sqrt{y}}\)<=>\(1=\frac{2}{2\sqrt{y}-\sqrt{y}}\)<=>\(1=\frac{2}{\sqrt{y}}\)<=> \(\sqrt{y}=2\)
<=> \(\left(\sqrt{y}\right)^2=2^2\)<=> \(y=4\)
=>\(x=4y=4\cdot4=16\)
a)
\(M=\frac{-(\sqrt{x}+1)\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-2\right)\left(x+2\right)}+\frac{-2\sqrt{x}\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)\left(x+2\right)}+\frac{2+5\sqrt{x}}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-x-3\sqrt{x}-2-2x+4\sqrt{x}+2+5\sqrt{x}}{4-x}\)
\(=\frac{-3x+6\sqrt{x}}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-3\sqrt{x}\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-3\sqrt{x}}{-\sqrt{x}-2}\)
a)\(M=\sqrt{x+\sqrt{x^2-4}}\sqrt{x-\sqrt{x^2-4}}\)
=\(\sqrt{\left(x+\sqrt{x^2-4}\right)\left(x-\sqrt{x^2-4}\right)}\)
=\(\sqrt{x^2-\left(\sqrt{x^2-4}\right)^2}\)
=\(\sqrt{x^2-\left(x^2-4\right)}\)
=\(\sqrt{x^2-x^2+4}\)
=\(\sqrt{4}=2\)
b) vì M=2 nên giá trị của M không phụ thuộc vào giá trị của biến nên với
\(x=4+\sqrt{5}\)
thì giá trị của M vẫn là 4
\(M\sqrt{x}=\sqrt{\left(x+2\right)+\left(x-2\right)+2\sqrt{\left(x-2\right)\left(x+2\right)}}+\sqrt{\left(x+2\right)+\left(x-2\right)-2\sqrt{\left(x-2\right)\left(x+2\right)}}\)
\(=\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{x+2}-\sqrt{x-2}\right)^2}\)
\(=\sqrt{x+2}+\sqrt{x-2}+\sqrt{x+2}-\sqrt{x-2}=2\sqrt{x+2}\)
\(\Rightarrow M=\sqrt{2}\sqrt{x+2}\)