Question

Tính  \(M=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\) Biểu thức \(M=\frac{\left(x-1\right)\sqrt{3}}{\sqrt{x^2-x+1}}\).Tính giá trị của M tại\(x=2+\sqrt{3}\)

Answer 1

1)

\(M=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{4+2.2.\sqrt{2}+2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{4-2.2.\sqrt{2}+2}}\)

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{\left(2+\sqrt{2}\right)^2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\)

\(=\frac{6+4\sqrt{2}}{2+2\sqrt{2}}+\frac{6-4\sqrt{2}}{-2+2\sqrt{2}}\)

\(=\frac{2.\left(3+2\sqrt{2}\right)}{2.\left(1+\sqrt{2}\right)}+\frac{2.\left(3-2\sqrt{2}\right)}{2.\left(\sqrt{2}-1\right)}\)

\(=\frac{3+2\sqrt{2}}{\sqrt{2}+1}+\frac{3-2\sqrt{2}}{\sqrt{2}-1}\)

\(=\frac{\left(3+2\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=1+\sqrt{2}+\sqrt{2}-1=2\sqrt{2}\)