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a: \(=12\sqrt{80}=48\sqrt{5}\)
b: \(=2\sqrt{5}\cdot2\sqrt{3}-10=4\sqrt{15}-10\)
c: =20-9=11
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)
\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)
\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)
3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)
=-2
4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
1)\(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\left|\sqrt{3}-1\right|=3-\sqrt{3}+\sqrt{3}-1=2\)
\(a,\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\)
\(=\sqrt{5}-2-\left(1+\sqrt{5}\right)\)
\(=\sqrt{5}-2-1-\sqrt{5}\)
\(=-3\)
\(b,\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+6\cdot\dfrac{\sqrt{4}}{\sqrt{3}}\)
\(=\sqrt{3}+\dfrac{12}{\sqrt{3}}\)
\(=\sqrt{3}+\dfrac{12\sqrt{3}}{3}\)
\(=\sqrt{3}+4\sqrt{3}\)
\(=5\sqrt{3}\)
#\(Toru\)
\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\\ =\sqrt{5}-2-1-\sqrt{5}\\ =-2-1\\ =-3\)
\(\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\\ =\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+4\sqrt{3}\\ =\sqrt{3}+4\sqrt{3}\\ =5\sqrt{3}\)
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=4
a) \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2\left(6+2\sqrt{5}\right)}{2}=\dfrac{\left(\sqrt{5}-1\right)^2\left(\sqrt{5}+1\right)^2}{2}=\dfrac{\left[\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\right]^2}{2}=\dfrac{\left(5-1\right)^2}{2}=8\)