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\(x=\frac{1}{2}\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\frac{1}{2}.\left(\sqrt{2}-1\right)\)
\(\Rightarrow2x=\sqrt{2}-1\Rightarrow2x+1=\sqrt{2}\)
\(\Rightarrow4x^2+4x+1=2\Rightarrow4x^2+4x-1=0\)
\(B=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1-1\right]^{2018}+2018\)
\(=\left(-1\right)^{2018}+2018=2019\)
Lời giải:
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Lời giải:
\(N=\sqrt{4\sqrt{6}+8\sqrt{3}+4\sqrt{2}+18}\)
\(=\sqrt{2\sqrt{24}+4(2\sqrt{3}+\sqrt{2})+18}\)
\(=\sqrt{12+2\sqrt{24}+2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2})^2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2}+2)^2}=\sqrt{12}+\sqrt{2}+2=2\sqrt{3}+\sqrt{2}+2\)
\(A=\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
=\(\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(=3-\sqrt{3}+2\sqrt{3}-3\)
=\(\sqrt{3}\)