Bài 1: Tính:
a) \(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right)\) : \(\sqrt{7}\)
b) \(\sqrt{36+12\sqrt{15}}\) : \(\sqrt{6}\)
c) \(\left(\sqrt{\dfrac{1}{3}-\sqrt{\dfrac{4}{3}}+\sqrt{3}}\right)\) : \(\sqrt{3}\)
Bài 2: Tính:
a) A=\(\left(\sqrt{27}-\sqrt{12}+2\sqrt{6}\right)\) : \(3\sqrt{3}\)
b) B=\(\left(\sqrt{12}-2\sqrt{18}\right)\) . \(\dfrac{\sqrt{2}}{2}\)
c) C=\(\left(\dfrac{1-\sqrt{2}}{1+\sqrt{2}}.\dfrac{1+\sqrt{2}}{1-\sqrt{2}}\right)\) : \(\sqrt{72}\)
Bài 1:
a: \(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
\(=\left(\dfrac{1}{7}\sqrt{7}-\dfrac{4}{7}\sqrt{7}+\sqrt{7}\right):\sqrt{7}\)
\(=\dfrac{1}{7}-\dfrac{4}{7}+1=1-\dfrac{3}{7}=\dfrac{4}{7}\)
b: \(\sqrt{36+12\sqrt{15}}:\sqrt{6}\)
\(=\sqrt{\dfrac{6\left(6+2\sqrt{15}\right)}{6}}=\sqrt{6+2\sqrt{15}}\)
c: Sửa đề: \(\left(\sqrt{\dfrac{1}{3}}-\sqrt{\dfrac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=\left(\dfrac{1}{3}\sqrt{3}-\dfrac{2}{3}\sqrt{3}+\sqrt{3}\right):\sqrt{3}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}+1=1-\dfrac{1}{3}=\dfrac{2}{3}\)
Bài 2:
a: \(A=\left(\sqrt{27}-\sqrt{12}+2\sqrt{6}\right):3\sqrt{3}\)
\(=\dfrac{\left(3\sqrt{3}-2\sqrt{3}+2\sqrt{6}\right)}{3\sqrt{3}}\)
\(=\dfrac{\sqrt{3}+2\sqrt{6}}{3\sqrt{3}}=\dfrac{1+2\sqrt{2}}{3}\)
b: \(B=\left(\sqrt{12}-2\sqrt{18}\right)\cdot\dfrac{\sqrt{2}}{2}\)
\(=\left(2\sqrt{3}-2\cdot3\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}\)
\(=\left(\sqrt{3}-3\sqrt{2}\right)\cdot\sqrt{2}=\sqrt{6}-6\)
