Ta có cái đầu <5

Cái sau <3 nên VT <8

Cảm ơn bạn nhe

Đặt \(\sqrt[3]{5\sqrt[]{2}+7}-\sqrt[3]{5\sqrt[]{2}-7}=x>0\)

\(\Rightarrow x^3=14-3\left(\sqrt[3]{5\sqrt[]{2}+7}-\sqrt[3]{5\sqrt[]{2}-7}\right)\sqrt[3]{\left(5\sqrt[]{2}+7\right)\left(5\sqrt[]{2}-7\right)}\)

\(\Rightarrow x^3=14-3x.\sqrt[3]{\left(5\sqrt[]{2}\right)^2-7^2}\)

\(\Rightarrow x^3=14-3x\)

\(\Rightarrow x^3+3x-14=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)=0\)

\(\Rightarrow x=2\)

\(\dfrac{4}{\sqrt{5}-1}+\dfrac{3}{\sqrt{5}-2}-\dfrac{16}{3-\sqrt{5}}\)

\(=\sqrt{5}+1+3\sqrt{5}+6-12-4\sqrt{5}\)

=-5

Đề bài đúng: \(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=1\)

Hoặc: \(\dfrac{\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}=1\)

\(=\dfrac{\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)

\(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=\dfrac{\sqrt{8-2\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{2}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)

Đặt \(a=\sqrt[4]{5}\Leftrightarrow5=a^4\)

Ta cần chứng minh: \(\left(\frac{a+1}{a-1}\right)^4=\frac{3+2a}{3-2a}\)

Khai triển: \(VT=\left(\frac{a+1}{a-1}\right)^4=\frac{\left(a+1\right)^4}{\left(a-1\right)^4}\)

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

                                         \(\frac{3+2a}{3-2a}=VP\)(đpcm)

DONE!

\(\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}+\sqrt{8}+\sqrt{10}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)

\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)+\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{4}+\sqrt{2}.\sqrt{5}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)

\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)\left(1+\sqrt{2}\right)}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)

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

⇒ ĐPCM