\(\hept{\begin{cases}\left(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}\right)^2=9\left(\sqrt[3]{5}-\sqrt[3]{4}\right)\\\left(3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}\right)^2=9\left(\sqrt[3]{5}-\sqrt[3]{4}\right)\end{cases}}\)

\(\left(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}\right)^2\)

\(=\sqrt[3]{4}+2\sqrt[3]{50}+5\sqrt[3]{5}+2\left(2\sqrt[3]{5}-\sqrt[3]{50}-5\sqrt[3]{4}\right)\)

\(=9\sqrt[3]{5}-9\sqrt[3]{4}=9\left(\sqrt[3]{5}-\sqrt[3]{4}\right)\)

\(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}=3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}\)

thanks bạn nhiều nha!!!! Chúc bạn hok tốt

Câu b, c tương tự câu a. Mình làm câu a coi như tượng trưng nha !!!!!!

a) Đặt: \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

<=> \(A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}.\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

<=> \(A^3=4+3\sqrt[3]{4-5}.A\)

<=> \(A^3=4-3A\)

<=> \(A^3+3A-4=0\)

<=> \(\left(A-1\right)\left(A^2+A+4\right)=0\)

Có:     \(A^2+A+4=\left(A+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

=>    \(A-1=0\)

<=> \(A=1\)

=> \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1\)

VẬY TA CÓ ĐPCM

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

b) \(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}=\dfrac{\sqrt{3}+1-\left(\sqrt{3}-1\right)}{3-1}=1\)

c) \(2\sqrt{5}-3\sqrt{45}+\sqrt{500}=2\sqrt{5}-9\sqrt{5}+10\sqrt{5}=3\sqrt{5}\)

d) \(\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\sqrt{3}=\dfrac{1-\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1-3-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-2-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=-\sqrt{2}\)

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

f) \(\sqrt{3}-\sqrt{4+2\sqrt{3}}=\sqrt{3}-\left(\sqrt{3}+1\right)=-1\)

g) \(\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\sqrt{5}-\dfrac{4}{\sqrt{5}+1}=\dfrac{5+\sqrt{5}-4}{\sqrt{5}+1}=1\)

h)\(\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}=\sqrt{37-20\sqrt{3}+\left(5+2\sqrt{3}\right)}=\sqrt{42-18\sqrt{3}}=\sqrt{\left(3\sqrt{3}+3\right)^2+6}\)

\(1,\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)

\(=\left(\sqrt{9.5}\sqrt{4.5}+\sqrt{5}\right).\frac{1}{\sqrt{6}}\)

\(=\frac{2\sqrt{5}}{\sqrt{6}}\)

\(=\frac{\sqrt{30}}{3}\)

1) \(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)

\(=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\)

\(=\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\)

\(=\frac{2\sqrt{5}}{\sqrt{6}}\)

\(=\frac{2\sqrt{5}\sqrt{6}}{\sqrt{6}.\sqrt{6}}\)

\(=\frac{2\sqrt{30}}{6}\)

\(=\frac{\sqrt{30}}{3}\)