Bài 1:
\(a,\dfrac{\sqrt{6}+2\sqrt{3}}{\sqrt{2}+1}-\sqrt{7+2\sqrt{6}}+1\\ =\dfrac{\sqrt{6}\left(1+\sqrt{2}\right)}{1+\sqrt{2}}-\sqrt{\left(\sqrt{6}+1\right)^2}+1\\ =\sqrt{6}-\sqrt{6}-1+1\\ =0\\ b,\left(\dfrac{\sqrt{7}+\sqrt{21}}{\sqrt{3}+1}+\sqrt{8-2\sqrt{7}}+1\right)\left(\sqrt{28}+\sqrt{7}\right)\\ =\left(\sqrt{7}+\sqrt{7}-1+1\right)\left(\sqrt{28}+\sqrt{7}\right)\\ =2\sqrt{7}\left(\sqrt{28}+\sqrt{7}\right)\\ =42\)
Bài 2:
a, Điều kiện: x\(\ge\)7
\(3\sqrt{x-7}-4=11\\ \Leftrightarrow3\sqrt{x-7}=15\\ \Leftrightarrow\sqrt{x-7}=5\\ \Leftrightarrow x-7=25\\ \Leftrightarrow x=32\left(tm\right)\)
b, Điều kiện: \(x\le2\)
\(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\\ \Leftrightarrow\dfrac{5}{2}\sqrt{2-x}-8\sqrt{2-x}+3\sqrt{2-x}=-10\\ \Leftrightarrow-\dfrac{5}{2}\sqrt{2-x}=-10\\ \Leftrightarrow\sqrt{2-x}=4\\ \Leftrightarrow x=-2\left(tm\right)\)
Bài 2:
c, Điều kiện: \(x\ge-5\)
\(\sqrt{2x^2-4x+25}-x=5\\ \Leftrightarrow2x^2-4x+25=x^2+10x+25\\ \Leftrightarrow x^2-14x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\)
d, Điều kiện: \(x\ge2\)
\(\sqrt{x+1}-\sqrt{x-2}=1\\ \Leftrightarrow\sqrt{x+1}=\sqrt{x-2}+1\\ \Leftrightarrow x+1=x-2+1+2\sqrt{x-2}\\ \Leftrightarrow2\sqrt{x-2}=2\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\\ \Leftrightarrow x=3\left(tm\right)\)
Bài 1:
a) \(\dfrac{\sqrt{6}+2\sqrt{3}}{\sqrt{2}+1}-\sqrt{7+2\sqrt{6}}+1\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{\left(\sqrt{6}\right)^2+2\cdot\sqrt{6}\cdot1+1^2}+1\)
\(=\sqrt{6}-\sqrt{\left(\sqrt{6}+1\right)^2}+1\)
\(=\sqrt{6}-\left|\sqrt{6}+1\right|+1\)
\(=\sqrt{6}-\sqrt{6}-1+1\)
\(=0\)
Bài 1:
b) \(\left(\dfrac{\sqrt{7}+\sqrt{21}}{\sqrt{3}+1}+\sqrt{8-2\sqrt{7}}+1\right)\left(\sqrt{28}+\sqrt{7}\right)\)
\(=\left[\dfrac{\sqrt{7}\left(1+\sqrt{3}\right)}{\sqrt{3}+1}+\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}+1\right]\left(2\sqrt{7}+\sqrt{7}\right)\)
\(=\left[\sqrt{7}+\sqrt{\left(\sqrt{7}-1\right)^2}+1\right]\cdot3\sqrt{7}\)
\(=\left(\sqrt{7}+\left|\sqrt{7}-1\right|+1\right)\cdot3\sqrt{7}\)
\(=\left(\sqrt{7}+\sqrt{7}-1+1\right)\cdot3\sqrt{7}\)
\(=2\sqrt{7}\cdot3\sqrt{7}\)
\(=6\cdot7\)
\(=42\)
