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a) \(\sqrt{\frac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}=\sqrt{\frac{1}{2}}+\sqrt{\frac{9}{2}}+\sqrt{\frac{25}{2}}=\sqrt{\frac{1}{2}}+3\sqrt{\frac{1}{2}}+5\sqrt{\frac{1}{2}}=9\sqrt{\frac{1}{2}}\)
b) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=\sqrt{4.5}-\sqrt{9.5}+3\sqrt{9.2}+\sqrt{36.2}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
a) \(\sqrt{\frac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}=\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}\)
b) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}=15\sqrt{2}-\sqrt{5}\)
a. \(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right).\sqrt{3}=\left(\left|\sqrt{3}-\sqrt{2}\right|+\sqrt{2}\right).\sqrt{3}=\left(\sqrt{3}\right)^2=3\)
b.\(\frac{2-\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}\left(2-\sqrt{2}\right)}{2}=\frac{2\sqrt{2}-2}{2}=\frac{2\left(\sqrt{2}-1\right)}{2}=\sqrt{2}-1\)
Bài 1:
a) Để căn thức \(\sqrt{\frac{2}{9-x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\frac{2}{9-x}\ge0\\9-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9-x>0\\x\ne9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 9\\x\ne9\end{matrix}\right.\Leftrightarrow x< 9\)
b) Ta có: \(x^2+2x+1\)
\(=\left(x+1\right)^2\)
mà \(\left(x+1\right)^2\ge0\forall x\)
nên \(x^2+2x+1\ge0\forall x\)
Do đó: Căn thức \(\sqrt{x^2+2x+1}\) xác được với mọi x
c) Để căn thức \(\sqrt{x^2-4x}\) có nghĩa thì \(x^2-4x\ge0\)
\(\Leftrightarrow x\left(x-4\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-4\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x< 0\end{matrix}\right.\)
Bài 3:
a) Ta có: \(\sqrt{\left(3-\sqrt{10}\right)^2}\)
\(=\left|3-\sqrt{10}\right|\)
\(=\sqrt{10}-3\)(Vì \(3< \sqrt{10}\))
b) Ta có: \(\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left|\sqrt{5}-2\right|\)
\(=\sqrt{5}-2\)(Vì \(\sqrt{5}>2\))
c) Ta có: \(3x-\sqrt{x^2-2x+1}\)
\(=3x-\sqrt{\left(x-1\right)^2}\)
\(=3x-\left|x-1\right|\)
\(=\left[{}\begin{matrix}3x-\left(x-1\right)\left(x\ge1\right)\\3x-\left(1-x\right)\left(x< 1\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}3x-x+1\\3x-1+x\end{matrix}\right.=\left[{}\begin{matrix}2x+1\\4x-1\end{matrix}\right.\)
\(a.\left(2-\sqrt{10}\right)\times\left(\sqrt2-\sqrt5\right)\)
\(=\sqrt2\cdot\left(\sqrt2-\sqrt5\right)\times\left(\sqrt2-\sqrt5\right)\)
\(=\sqrt2\cdot\left(\sqrt2-\sqrt5\right)^2=\sqrt2\cdot\left(2-2\sqrt{10}+5\right)\)
\(=\sqrt2\cdot\left(7-2\sqrt{10}\right)=7\sqrt2-2\sqrt{20}\)
\(=7\sqrt2-4\sqrt5\)
\(b.\sqrt3\times\left(\sqrt{72}+\sqrt{4,5}-\sqrt{12,5}\right)\)
\(=\sqrt3\times\left(6\sqrt2+\frac{3\sqrt2}{2}-\frac{5\sqrt2}{2}\right)\)
\(=6\sqrt6+\frac32\sqrt6-\frac52\sqrt6\)
\(=\sqrt6\times\left(6+\frac32-\frac52\right)\)
\(=\sqrt6\times5=5\sqrt6\)
\(c.12\times\left(\sqrt{\frac23}-\sqrt{\frac32}\right)=12\times\left(\frac{\sqrt2}{\sqrt3}-\frac{\sqrt3}{\sqrt2}\right)\)
\(=12\times\left(\frac{\sqrt6}{3}-\frac{\sqrt6}{2}\right)=12\sqrt6\times\left(\frac13-\frac12\right)\)
\(=12\sqrt6\times\left(-\frac16\right)=-2\sqrt6\)