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\(A=43+24\sqrt{3}-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)
\(=43+24\sqrt{3}-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)
\(=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)
\(=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)\)
\(=43-8=35\)
\(A=\dfrac{2\left(\sqrt{2}+\sqrt{6}\right)}{3\sqrt{2+\sqrt{3}}}=\dfrac{4\left(\sqrt{2}+\sqrt{6}\right)}{3.\sqrt{4\left(2+\sqrt{3}\right)}}=\dfrac{4\left(\sqrt{2}+\sqrt{6}\right)}{3.\sqrt{8+2\sqrt{12}}}=\dfrac{4\left(\sqrt{2}+\sqrt{6}\right)}{3.\sqrt{\left(\sqrt{2}+\sqrt{6}\right)^2}}=\dfrac{4}{3}\)
\(B=\sqrt{8-2\sqrt{15}}-\sqrt{\left(3-3\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-3\sqrt{5}+3\\ =\sqrt{5}-\sqrt{3}-3\sqrt{5}+3=3-\sqrt{3}-2\sqrt{5}\)
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câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
\(A=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\) \(A=\left(6+7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(16+2.4.3\sqrt{3}+27\right)}}\)
\(A=6\left(7+4\sqrt{3}\right)+\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)Trong căn là hằng đẳng thức (a+b)^2
\(A=42+24\sqrt{3}+7^2-\left(4\sqrt{3}\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\) sử dụng hằng đẳng thức a^2 -b^2\(A=43+24\sqrt{3}-8\sqrt{20+8+2.3\sqrt{3}}\)
\(A=43+24\sqrt{3}-8\sqrt{1+2.3\sqrt{3}+27}\)trong căn tiếp tục là hằng đẳng thức (a+b)^2\(A=43+24\sqrt{3}-8\sqrt{\left(1+3\sqrt{3}\right)^2}\)
\(A=43+24\sqrt{3}-8\left(1+3\sqrt{3}\right)\)
\(A=35\)
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