Bài 1:
a) \(\sqrt{4\cdot\dfrac{1}{2}}+\sqrt{32}-\sqrt{72}+\sqrt{162}\)
\(=\sqrt{\dfrac{4}{2}}+\sqrt{4^2\cdot2}-\sqrt{6^2\cdot2}+\sqrt{9^2\cdot2}\)
\(=\sqrt{2}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\)
\(=8\sqrt{2}\)
b) \(\dfrac{7+2\sqrt{12}}{\sqrt{3}+\sqrt{4}}-\dfrac{7-2\sqrt{12}}{\sqrt{4}-\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}}{\sqrt{3}+\sqrt{4}}-\dfrac{7-4\sqrt{12}}{\sqrt{4}-\sqrt{3}}\)
\(=\dfrac{2^2+2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}{\sqrt{3}+2}-\dfrac{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}{2-\sqrt{3}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)^2}{2+\sqrt{3}}-\dfrac{\left(2-\sqrt{3}\right)^2}{2-\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c) \(\left(\dfrac{3}{\sqrt{2}+1}+\dfrac{14}{2\sqrt{2}-1}-\dfrac{4}{2-\sqrt{2}}\right)\cdot\left(\sqrt{8}+2\right)\)
\(=\left[\dfrac{3\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\dfrac{14\left(2\sqrt{2}+1\right)}{\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)}-\dfrac{4\left(2+\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\right]\cdot\left(2\sqrt{2}+2\right)\)
\(=\left[\dfrac{3\left(\sqrt{2}-1\right)}{2-1}+\dfrac{14\left(2\sqrt{2}+1\right)}{8-1}-\dfrac{4\left(2+\sqrt{2}\right)}{4-2}\right]\cdot\left(2\sqrt{2}+2\right)\)
\(=\left[3\left(\sqrt{2}-1\right)+2\left(2\sqrt{2}+1\right)-2\left(2+\sqrt{2}\right)\right]\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
\(=10\cdot\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=10\cdot\left(2-1\right)\)
\(=10\)
1:
a: \(\sqrt{4\dfrac{1}{2}}+\sqrt{32}-\sqrt{72}+\sqrt{162}\)
\(=\sqrt{\dfrac{9}{2}}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\)
\(=\dfrac{3}{2}\sqrt{2}+7\sqrt{2}=\dfrac{17}{2}\sqrt{2}\)
2:
a: \(\Leftrightarrow x+7-\sqrt{x+7}=0\)
=>\(\sqrt{x+7}\left(\sqrt{x+7}-1\right)=0\)
=>x+7=0 hoặc x+7=1
=>x=-6 hoặc x=-7
b: \(\Leftrightarrow2x-1-\sqrt{2x+19}=0\)
=>\(2x+19-\sqrt{2x+19}-20=0\)
=>\(\left(\sqrt{2x+19}-5\right)\left(\sqrt{2x+19}+4\right)=0\)
=>2x+19=25
=>2x=6
=>x=3
c: \(\Leftrightarrow3\sqrt{x+1}\cdot\dfrac{1}{3}-2\sqrt{x+1}+8\cdot\dfrac{2}{5}\sqrt{x+1}=11\)
=>\(\dfrac{11}{5}\sqrt{x+1}=11\)
=>\(\sqrt{x+1}=5\)
=>x+1=25
=>x=24
