Question

a)\(\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\left(9x+5\right)\)

b)\(\sqrt{9-4\sqrt{5}}\)

c)\(\sqrt{9-6x+4x^2}\)=4

d)\(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\)

Answer 1

b)\(\sqrt{9-4\sqrt{5}}\)=\(\sqrt{9-\sqrt{80}}\)=\(\sqrt{\dfrac{9+\sqrt{9^2-80}}{2}}-\sqrt{\dfrac{9-\sqrt{9^2-80}}{2}}\)=\(\sqrt{5}\)\(-\)\(\sqrt{4}\)=\(2-\sqrt{5}\)

(dựa theo công thức có sẵn từ một quyển sách nâng cao:\(\sqrt{A\pm\sqrt{B}}\)=\(\sqrt{\dfrac{A+\sqrt{A^2-B}}{2}}\pm\sqrt{\dfrac{A-\sqrt{A^2-B}}{2}}\)

Answer 2

c: \(\Leftrightarrow4x^2-6x+9=16\)

\(\Leftrightarrow4x^2-6x-7=0\)

hay \(x\in\left\{\dfrac{3+\sqrt{37}}{4};\dfrac{3-\sqrt{37}}{4}\right\}\)

d: \(=\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\)

\(=\dfrac{1}{2}\sqrt{3}+\dfrac{5}{2}\)