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Lời giải:
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
\(a.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\sqrt{2}+30}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(b,\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)
\(=\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)
\(=\sqrt{m-1}+1+|\sqrt{m-1}-1|\)
\(A=\left(\dfrac{6x+4}{3\sqrt{3x^3}-8}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right).\left(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
Điều kiện tự làm nha:
Đặt \(\sqrt{3x}=a\) thì ta có:
\(A=\left(\dfrac{2a^2+4}{a^3-8}-\dfrac{a}{a^2+2a+4}\right).\left(\dfrac{1+a^3}{1+a}-a\right)\)
\(=\left(\dfrac{2a^2+4}{\left(a-2\right)\left(a^2+2a+4\right)}-\dfrac{a}{a^2+2a+4}\right).\left(\dfrac{\left(1+a\right)\left(1-a+a^2\right)}{1+a}-a\right)\)
\(=\dfrac{a^2+2a+4}{\left(a-2\right)\left(a^2+2a+4\right)}.\left(1-2a+a^2\right)\)
\(=\dfrac{\left(a-1\right)^2}{a-2}=\dfrac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\)
Ta có \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\left(\sqrt{10}-\sqrt{2}\right)\)
= \(2\sqrt{4+\sqrt{\sqrt{5}^2-2\sqrt{5}.1+1}}\sqrt{2}\left(\sqrt{5}-1\right)\)
= \(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\sqrt{2}\left(\sqrt{5}-1\right)\)
= \(\sqrt{2}\sqrt{4+\sqrt{5}-1}.\left(\sqrt{5}-1\right)2\)
= \(\sqrt{2\left(3+\sqrt{5}\right)}\left(\sqrt{5}-1\right)2\)
= \(\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)2\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)2\)
= \(\left(\sqrt{5}^2-1\right)2\)
= 4.2
= 8
Chúc bạn làm bài tốt :)
Lời giải:
\(N=\sqrt{4\sqrt{6}+8\sqrt{3}+4\sqrt{2}+18}\)
\(=\sqrt{2\sqrt{24}+4(2\sqrt{3}+\sqrt{2})+18}\)
\(=\sqrt{12+2\sqrt{24}+2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2})^2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2}+2)^2}=\sqrt{12}+\sqrt{2}+2=2\sqrt{3}+\sqrt{2}+2\)