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a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
Bài 2:
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)
\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)
\(=10\sqrt{5}+6\sqrt{6}-6\sqrt{5}-2\sqrt{5}\)
\(=8\sqrt{5}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2}-\sqrt{\left(\sqrt{7}\right)^2+2\cdot3\cdot\sqrt{7}+3^2}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(\sqrt{7}+3\right)^2}\)
\(=\sqrt{7}-2-\sqrt{7}-3\)
\(=-5\)
\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
a) Ta có: \(\sqrt{11+6\sqrt{2}}-\left(3+\sqrt{2}\right)\)
\(=3+\sqrt{2}-3-\sqrt{2}\)
=0
b) Ta có: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\sqrt{2}+1\)
e) \(\sqrt{x^2}=\left|-8\right|\Rightarrow\left|x\right|=8\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
e) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}=\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{\dfrac{8+2\sqrt{7}}{2}}+\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}+\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}+\sqrt{2}\)
\(=\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}+\sqrt{2}=\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{\sqrt{7}+1}{\sqrt{2}}+\sqrt{2}\)
\(=-\dfrac{2}{\sqrt{2}}+\sqrt{2}=-\sqrt{2}+\sqrt{2}=0\)
f) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{12+2\sqrt{11}}{2}}-\sqrt{\dfrac{12-2\sqrt{11}}{2}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}\right)^2+2.\sqrt{11}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}\right)^2-2.\sqrt{11}.1+1^2}{2}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}-1\right)^2}{2}}+3\sqrt{2}\)
\(=\dfrac{\left|\sqrt{11}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{11}-1\right|}{\sqrt{2}}+3\sqrt{2}=\dfrac{\sqrt{11}+1}{\sqrt{2}}-\dfrac{\sqrt{11}-1}{\sqrt{2}}+3\sqrt{2}\)
\(=\dfrac{2}{\sqrt{2}}+3\sqrt{2}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\)