e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

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

=2

\(\left\{{}\begin{matrix}2\left(\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}\right)=\sqrt[4]{8\left(x^4+y^4\right)}+2\sqrt{xy}\left(1\right)\\16x^5-20x^3+5\sqrt{xy}=\sqrt{\dfrac{y+1}{2}}\left(2\right)\end{matrix}\right.\).

ĐKXĐ: \(xy>0;y\ge-\dfrac{1}{2}\).

Nhận thấy nếu x < 0 thì y < 0. Suy ra VT của (1) âm, còn VP của (1) dương (vô lí)

Do đó x > 0 nên y > 0.

Với a, b > 0 ta có bất đẳng thức \(\left(a+b\right)^4\le8\left(a^4+b^4\right)\).

Thật vậy, áp dụng bất đẳng thức Cauchy - Schwarz ta có:

\(\left(a+b\right)^4\le\left[2\left(a^2+b^2\right)\right]^2=4\left(a^2+b^2\right)^2\le8\left(a^4+b^4\right)\).

Dấu "=" xảy ra khi và chỉ khi a = b.

Áp dụng bất đẳng thức trên ta có:

\(\left(\sqrt[4]{8\left(x^4+y^4\right)}+2\sqrt{xy}\right)^4\le8\left[8\left(x^4+y^4\right)+16x^2y^2\right]=64\left(x^2+y^2\right)^2\)

\(\Rightarrow\left(\sqrt[4]{8\left(x^4+y^4\right)}+2\sqrt{xy}\right)^2\le8\left(x^2+y^2\right)\). (3)

Lại có \(4\left(\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}\right)^2=4\left(\dfrac{x^6}{y^4}+2xy+\dfrac{y^6}{x^4}\right)\). (4) 

Áp dụng bất đẳng thức AM - GM ta có \(\dfrac{x^6}{y^4}+xy+xy+xy+xy\ge5x^2;\dfrac{y^6}{x^4}+xy+xy+xy+xy\ge5y^2;3\left(x^2+y^2\right)\ge6xy\).

Cộng vế với vế của các bđt trên lại rồi tút gọn ta được \(\dfrac{x^6}{y^4}+2xy+\dfrac{y^6}{x^4}\ge2\left(x^2+y^2\right)\). (5)

Từ (3), (4), (5) suy ra \(4\left(\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}\right)^2\ge\left(\sqrt[4]{8\left(x^4+y^4\right)}+2\sqrt{xy}\right)^2\Rightarrow2\left(\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}\right)\ge\sqrt[4]{8\left(x^4+y^4\right)}+2\sqrt{xy}\).

Do đó đẳng thức ở (1) xảy ra nên ta phải có x = y.

Thay x = y vào (2) ta được:

\(16x^5-20x^3+5x=\sqrt{\dfrac{x+1}{2}}\). (ĐK: \(x>0\))

PT này có một nghiệm là x = 1 mà sau đó không biết giải ntn :v

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

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

\(=\sqrt{\left(9-5\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(9-5\right)\left(3+\sqrt{5}\right)}\)

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

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

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

\(=6+2\sqrt{9-5}=6+2\sqrt{4}=6+2\cdot2=6+4=10\)

\(\Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{10}\Rightarrow2\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)=2\sqrt{10}\)

\(\Rightarrow\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}=2\sqrt{10}\)

a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)

\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\left(\sqrt{10}+\sqrt{2}\right)\)

\(=\left(9-5\right).\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{5-2\sqrt{5}+1}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\)

\(=4.\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4.\left(5-1\right)=16\)

b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{5-2\sqrt{5}+1}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\)

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

\(=2\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)

\(=2\sqrt{5+2\sqrt{5}+1}.\left(\sqrt{5}-1\right)\)

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

\(=2.\left(5-1\right)=2.4=8\)

Bài 20:

a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)

b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)

\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)

c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

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

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

=2

d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

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

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

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

=2

b,\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)  \(=\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}+4\sqrt{8\sqrt{3}}\)

\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}\)

\(=0\)

d,\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(\sqrt{2}A=\sqrt{2}(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})\)

\(\sqrt2A=\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

\(\sqrt2A=\sqrt{(\sqrt5-1)^2}\) \(+\sqrt{(\sqrt5+1)^2}\)    \(=\sqrt5-1 +\sqrt5+1=2\sqrt5\)

\(\Rightarrow A=\dfrac{2\sqrt5}{\sqrt2}\) \(=\sqrt{10}\)

a. \(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)

\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\) 

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

\(=\frac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=\frac{2\left(\sqrt{5}+1\right)}{2\left(\sqrt{5}+1\right)}=1\)