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a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
1.
\(\sqrt{14+6\sqrt{5}}-\sqrt{\dfrac{\sqrt{5}-2}{\sqrt{5}+2}}\)
=\(\sqrt{9+6\sqrt{5}+5}-\dfrac{\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+2}}\)
=\(\sqrt{\left(3+\sqrt{5}\right)^2}-\dfrac{\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}}{\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)}}\)
= \(3+\sqrt{5}-\dfrac{\sqrt{5-4}}{\sqrt{\left(\sqrt{5}+2\right)^2}}\)
= \(\dfrac{3\left(\sqrt{5}+2\right)}{\sqrt{5+2}}+\dfrac{\sqrt{5}\left(\sqrt{5}+2\right)}{\sqrt{5}+2}-\dfrac{1}{\sqrt{5}+2}\)
=\(\dfrac{5\sqrt{5}+10}{\sqrt{5}+2}=\dfrac{5\left(\sqrt{5}+2\right)}{\sqrt{5}+2}=5\)
2, \(\sqrt{4x+8}+\sqrt{9x+18}-\sqrt{9}=\sqrt{16x+32}\)
⇔\(\sqrt{4\left(x+2\right)}+\sqrt{9\left(x+2\right)}-3=\sqrt{16\left(x+2\right)}\)
⇔\(2\sqrt{x+2}+3\sqrt{x+2}-4\sqrt{x+2}=3\)
\(\Leftrightarrow\sqrt{x+2}=3\)
⇔\(x+2=9\)
⇔x=7
Lời giải:
a) Ta có:
\(14-6\sqrt{5}=14-2\sqrt{45}=9+5-2\sqrt{9.5}=(\sqrt{9}-\sqrt{5})^2=(3-\sqrt{5})^2\)
\(\Rightarrow \sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)
\(6+2\sqrt{5}=5+1+2\sqrt{5.1}=(\sqrt{5}+1)^2\)
\(\Rightarrow \sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
Do đó: \(\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}=3-\sqrt{5}+\sqrt{5}+1=4\)
b)
\(\frac{\sqrt{10}+10}{1+\sqrt{10}}-\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{10}(1+\sqrt{10})}{1+\sqrt{10}}-\frac{\sqrt{10}(\sqrt{5}-\sqrt{2})}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{10}-\sqrt{10}=0\)
Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)
c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)
e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
1.
a, \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}=6\sqrt{2}-20\sqrt{2}-12\sqrt{2}=-2\sqrt{2}\)
b, \(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}+3\right)^2}\)
\(=\left|\sqrt{5}-3\right|+\left|\sqrt{5}+3\right|\)
\(=-\sqrt{5}+3+\sqrt{5}+3=6\)
c, \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\dfrac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{10}-\sqrt{10}=0\)
2.
ĐK: \(x\in R\)
\(\sqrt{9x^2-30x+25}=5\)
\(\Leftrightarrow\sqrt{\left(3x-5\right)^2}=5\)
\(\Leftrightarrow\left|3x-5\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=5\\3x-5=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=0\end{matrix}\right.\)
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