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Bài 2 :
Ta có : \(\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(5+3-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left(16-15\right)=2.1=2\)
Bài 1 :
a, ĐKXĐ : \(x\ge0\)
Ta có : \(PT\Leftrightarrow3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21\)
\(\Leftrightarrow7\sqrt{5x}=21\)
\(\Leftrightarrow\sqrt{5x}=3\)
\(\Leftrightarrow x=\dfrac{9}{5}\left(TM\right)\)
Vậy ...
b, Ta có : \(PT\Leftrightarrow\sqrt{\left(x-5\right)^2}=4\)
\(\Leftrightarrow\left|x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy ....
\(=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{6^2}}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5+2\sqrt{3}\sqrt{5}+3}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)
\(VT\Leftrightarrow\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2=VP\left(dpcm\right)\)
a, \(VT=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{20}-2\right)}{2}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}\left(3+\sqrt{5}\right)\left(2\sqrt{5}-2\right)}{2}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)2\left(\sqrt{5}-1\right)}{2}\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-10=8=VP\)
b, \(VT=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left(16-15\right)=2=VP\)
Sửa đề: Chứng minh \(\left(\sqrt{7+4\sqrt{3}}+\sqrt{8-2\sqrt{15}}\right)-\left(\sqrt{8+2\sqrt{15}}-\sqrt{7-4\sqrt{3}}\right)=\left(\sqrt{3}-1\right)^2\)
Ta có: \(VT=\left(\sqrt{7+4\sqrt{3}}+\sqrt{8-2\sqrt{15}}\right)-\left(\sqrt{8+2\sqrt{15}}-\sqrt{7-4\sqrt{3}}\right)\)
\(=\left(\sqrt{4+2\cdot2\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\right)-\left(\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\right)\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|-\left|\sqrt{5}+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)
\(=\left(2+\sqrt{3}\right)+\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}+\sqrt{3}\right)+\left(2-\sqrt{3}\right)\)
\(=2+\sqrt{3}+\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2-\sqrt{3}\)
\(=4-2\sqrt{3}\)
\(=3-2\cdot\sqrt{3}\cdot1+1\)
\(=\left(\sqrt{3}-1\right)^2=VP\)(đpcm)
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
Lời giải:
$(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}$
$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}$
$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}$
$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})$
$=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})$
$=2(4^2-15)=2$ (đpcm)