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1: \(=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
2: \(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
a ) \(\sqrt{3+2\sqrt[]{2}}\) - \(\sqrt{2}\)
= \(\sqrt{\left(1+\sqrt{2}\right)^2}\) -\(\sqrt{2}\)
= 1 + \(\sqrt{2}\) - \(\sqrt{2}\)
=1
b) \(\sqrt{16-6\sqrt{7}}\)-\(2\sqrt{7}\)
= \(\sqrt{\left(3-\sqrt{7}\right)^2}\)-\(2\sqrt{7}\)
= 3 - \(\sqrt{7}\)-\(2\sqrt{7}\)
=3 - 3\(\sqrt{7}\)
c )\(\sqrt{30+12\sqrt{6}}\) +\(\sqrt{30-12\sqrt{6}}\)
= \(\sqrt{6\left(5+2\sqrt{6}\right)}\) + \(\sqrt{6\left(5-2\sqrt{6}\right)}\)
=\(\sqrt{6}\) (\(\sqrt{5+2\sqrt{6}}\) + \(\sqrt{5-2\sqrt{6}}\) )
=\(\sqrt{6}\) [\(\sqrt{\left(1+\sqrt{6}\right)^2}\) +\(\sqrt{\left(1-\sqrt{6}\right)^2}\)
=\(\sqrt{6}\) (1 + \(\sqrt{6}\) + \(\sqrt{6}\) -\(1\))
= 2 . 6
=12
d)\(\sqrt{9-4\sqrt{5}}\) -\(\sqrt{5}\)
=\(\sqrt{\left(2-\sqrt{5}\right)}^2\) -\(\sqrt{5}\)
=\(\sqrt{5}\) -\(2\) -\(\sqrt{5}\)
=2
e ) \(\sqrt{\left(-2\right)^6}\) \(+\) \(\sqrt{\left(-3\right)}^4\)
= \(\left|\left(-2\right)^3\right|\) + \(\left|\left(-3\right)^2\right|\)
=8 + 9
=17
c)
\(\sqrt{2}C=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\)
\(=\sqrt{5}+1-\left(\sqrt{5}-1\right)-2=0\Rightarrow C=0\)
b)
\(B=3\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)\)
\(\Rightarrow\sqrt{2}B=3\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\)
\(=3\left(\sqrt{5}+1+\sqrt{5}-1\right)-\sqrt{5}\left(\sqrt{5}+1-\sqrt{5}+1\right)\)
\(\sqrt{2}B=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\Rightarrow B=2\sqrt{10}\)
C)√3+√5−√3−√5−√2b) (3−√5)√3+√5+(3+√5)√3−√5d) √4−√7−√4+√7+√7e) √6,5+√12+√6,5−√12+2√6mình cần giải gấp ạ
\(D^2=6\Rightarrow \left[\begin{matrix} D=\sqrt{6}\\ D=-\sqrt{6}\end{matrix}\right.\)
Mà $D< 0$ thì đương nhiên $D=-\sqrt{6}$ rồi em.
Em cảm ơn chị rất nhiều!