\(a,\left(3-\sqrt{7}\right)^2=3^2-2.3.\sqrt{7}+\sqrt{7^2}=9+7-6\sqrt{7}=16-6\sqrt{7}\left(dpcm\right)\)
\(b,\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{\sqrt{11^2}-2.3\sqrt{11}+9}=\sqrt{11}-\sqrt{\left(\sqrt{11}-3\right)^2}=\sqrt{11}-\left|\sqrt{11}-3\right|=\sqrt{11}-\left(\sqrt{11}-3\right)=3\left(dpcm\right)\)
\(c,\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}\)
\(=\sqrt{6^2+2.6\sqrt{5}+\sqrt{5}^2}-\sqrt{6^2-2.6\sqrt{5}+\sqrt{5^2}}\)
\(=\sqrt{\left(6+\sqrt{5}\right)^2}-\sqrt{\left(6-\sqrt{5}\right)^2}\)
\(=\left|6+\sqrt{5}\right|-\left|6-\sqrt{5}\right|\)
\(=6+\sqrt{5}-6+\sqrt{5}\)
\(=2\sqrt{5}\left(dpcm\right)\)
3)
a.
\(\left(3-\sqrt{7}\right)^2=3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2=9-6\sqrt{7}+7=16-6\sqrt{7}=VP\)
b.
\(\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{3^2-2.3.\sqrt{11}+11}=\sqrt{11}-\sqrt{\left(3-\sqrt{11}\right)^2}=\sqrt{11}-\left(\sqrt{11}-3\right)=3=VP\)
c.
\(\sqrt{\sqrt{5}^2+2.6.\sqrt{5}+6^2}-\sqrt{\sqrt{5}^2-2.6.\sqrt{5}+6^2}=\sqrt{\left(\sqrt{5}+6\right)^2}-\sqrt{\left(\sqrt{5}-6\right)^2}\\ =\sqrt{5}+6-\left(6-\sqrt{5}\right)=2\sqrt{5}=VP\)
4)
a. không thấy x
b. Để căn thức có nghĩa thì:
\(\left\{{}\begin{matrix}\dfrac{10-30x}{3x^2+1}\ge0\\3x^2+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10-30x\ge0\left(do.3x^2+1>0\forall x\right)\\x^2\ne-\dfrac{1}{3}\left(luôn.đúng\right)\end{matrix}\right.\\ \Rightarrow x\le\dfrac{1}{3}\)
c. Để căn thức có nghĩa thì:
\(\left\{{}\begin{matrix}\dfrac{4x+2}{x^2+4x+5}\ge0\\x^2+4x+5\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)^2+1>0\forall x\\4x+2\ge0\end{matrix}\right.\\ \Rightarrow x\ge-\dfrac{1}{2}\)
