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a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
a ) \(\dfrac{2}{\sqrt{3}-1}\) - \(\dfrac{2}{\sqrt{3}+1}\) = \(\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
= \(\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\) = \(\dfrac{4}{2}\) = 2
b) \(\dfrac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}\) - \(\dfrac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{5\left(2\sqrt{5}-3\sqrt{2}\right)-5\left(2\sqrt{5}+3\sqrt{2}\right)}{12\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12\left(20-18\right)}\)
= \(\dfrac{-30\sqrt{2}}{24}\) = \(\dfrac{-15\sqrt{2}}{12}\) = \(\dfrac{-5\sqrt{2}}{4}\)
c) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\) = \(\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
= \(\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\) = \(\dfrac{60}{20}\) = 3
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)
= \(\dfrac{\sqrt{3}}{\sqrt{2}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
a: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
b: \(=\sqrt{2}-1-\sqrt{2}=-1\)
\(a.\dfrac{\sqrt{8}+2}{\sqrt{2}+1}.\sqrt{2}=\dfrac{\sqrt{2}.2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\sqrt{2}\)
\(b.\dfrac{5+\sqrt{5}}{\sqrt{5}+1}.\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}.\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\sqrt{5}.\sqrt{5}=5\)
\(c.\dfrac{2}{\sqrt{2}}+\sqrt{18}+\sqrt{32}=\sqrt{2}+\sqrt{9.2}+\sqrt{16.2}=\sqrt{2}+3\sqrt{2}+4\sqrt{2}=8\sqrt{2}\)
\(d.\left(1-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\right)\left(1+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\right)=\left(1-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(1+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\right)=\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)=1-2=-1\)
\(A=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1-2\sqrt{x-1}+1}}=\dfrac{1-\sqrt{x-1}}{\sqrt{\left(\sqrt{x-1}-1\right)^2}}\)
\(A=\dfrac{1-\sqrt{x-1}}{\left|\sqrt{x-1}-1\right|}\) \(\Leftrightarrow\left[{}\begin{matrix}A=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1}-1}=-1\\A=\dfrac{1-\sqrt{x-1}}{-\sqrt{x-1}+1}=1\end{matrix}\right.\)
\(B=\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(B^2=\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)^2\)
\(B^2=8+2\sqrt{10+2\sqrt{5}}-2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}+8-2\sqrt{10+2\sqrt{5}}\)
\(B^2=16-2\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
\(B^2=16-2\sqrt{8^2-4\left(10+2\sqrt{5}\right)}\)
\(B^2=16-2\sqrt{24-8\sqrt{5}}\)
\(B^2=16-2\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(B^2=16-4\sqrt{5}+4=20-4\sqrt{5}\)
\(B=\sqrt{20-4\sqrt{5}}\)
\(a.B=\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
\(b.A=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\dfrac{1}{\sqrt{2}+1}-\sqrt{2}=\dfrac{-1-\sqrt{2}}{\sqrt{2}+1}=-1\)