x₀² = 8 - ( \(2\sqrt{2+\sqrt{3}}\)+ \(2\sqrt{6-3\sqrt{3}}\)) (1)

Ta có (  \(2\sqrt{2+\sqrt{3}}\)+ \(2\sqrt{6-3\sqrt{3}}\))² = 32

Do đó x₀² = 8 - \(\sqrt{32}\)(2)

PT <=> (x² - 8)² - 32 = 0 (3)

Thế (2) vào (3) thì đúng

Vậy x₀ là nghiệm của PT

Đặt \(x^2=t\left(t\ge0\right)\)

\(\Leftrightarrow t^2-16t+32=0\)

\(\Delta=\left(-16\right)^2-4.32=256-128=128>0\)

\(t_1=\frac{16-\sqrt{128}}{2}=8-4\sqrt{2};t_2=\frac{16+\sqrt{128}}{2}=8+4\sqrt{2}\)

Theo bài ra ta có : 

\(x_0=\sqrt{2+\sqrt{2+\sqrt{3}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\)

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

tịt lun, cái pt căn này chill quá 

 ๖²⁴ʱ๖ۣۜTɦủү❄吻༉ Mơn Bạn nha .

P/s : làm nháp thử mn sửa giúp nha ( thực ra em cũng chả hiểu cái gì cả T_T )

Ta có :

\(\left(x_0\right)^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3\left(2-\sqrt{3}\right)}\)

\(\Rightarrow\left(\frac{8-\left(x_0\right)^2}{2}\right)^2=2+\sqrt{3}+3\left(2-\sqrt{3}\right)+2\sqrt{3\left(4-3\right)}=8\)

\(\Rightarrow64-16\left(x_0\right)^2+\left(x_0\right)^4=32\)

\(\Rightarrow\left(x_0\right)^4-16\left(x_0\right)^2+32=0\left(đpcm\right)\)

ta có \(8-2\sqrt{3}+2\sqrt{3}=\left(2\cdot\sqrt{2}\right)^2\)

\(\Leftrightarrow\left(2+\sqrt{3}\right)+\left(6-3\sqrt{3}\right)+2\sqrt{\left(2+\sqrt{3}\right)\left(6-3\sqrt{3}\right)}=\left(2\cdot\sqrt{2}\right)^2\)

\(\Leftrightarrow\left(\sqrt{2+\sqrt{3}}+\sqrt{6-3\sqrt{3}}\right)^2=\left(2\cdot\sqrt{2}\right)^2\)

\(\Leftrightarrow\sqrt{2+\sqrt{3}}+\sqrt{6-3\sqrt{3}}=2\sqrt{2}\)

\(\Leftrightarrow8-2\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-\sqrt{2+\sqrt{3}}\right)}=8-4\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{2+\sqrt{2+\sqrt{3}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\right)^2=8-4\sqrt{2}\)

\(\Leftrightarrow x_0^2=8-4\sqrt{2}\)

\(\Leftrightarrow x_0^2-\left(8-4\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[x_0^2-\left(8-4\sqrt{2}\right)\right]\left[x_0^2-\left(8+4\sqrt{2}\right)\right]=0\)

\(\Leftrightarrow x_0^4-16x_0^2+32=0\)

\(x_0=\sqrt{2+\sqrt{2+\sqrt{3}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\)(x₀>0)

=> \(\left(x_0\right)^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{6-3\sqrt{2+\sqrt{3}}}\)

<=> \(\left(x_0\right)^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\)

<=> \(\left(x_0\right)^2=8-\sqrt{2}\sqrt{4+2\sqrt{3}}-2\sqrt{12-6\sqrt{2+\sqrt{3}}+6\sqrt{2+\sqrt{3}}-3\left(2+\sqrt{3}\right)}\)

<=> \(\left(x_0\right)^2=8-\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}-2\sqrt{12-6-3\sqrt{3}}=8-\sqrt{2}\left(\sqrt{3}+1\right)-2\sqrt{6-3\sqrt{3}}=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{2}\sqrt{12-6\sqrt{3}}\)

<=> \(\left(x_0\right)^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\sqrt{\left(3-\sqrt{3}\right)^2}=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\left|3-\sqrt{3}\right|=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\left(3-\sqrt{3}\right)\)

<=> \(\left(x_0\right)^2=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)

Có \(x^4-16x^2+32=0\) <=> \(x^4-8x^2+4\sqrt{2}x^2-8x^2+64-32\sqrt{2}-4\sqrt{2}x^2+32\sqrt{2}-32=0\)

<=> \(x^2\left(x^2-8+4\sqrt{2}\right)-8\left(x^2-8+4\sqrt{2}\right)-4\sqrt{2}\left(x^2-8+4\sqrt{2}\right)=0\)

<=>\(\left(x^2-8-4\sqrt{2}\right)\left(x^2-8+4\sqrt{2}\right)=0\)

=> \(\left[{}\begin{matrix}\left(x_1\right)^2=8+4\sqrt{2}\\\left(x_2\right)^2=8-4\sqrt{2}\end{matrix}\right.\) (x₁,x₂>0)

=> \(\left(x_0\right)^2=\left(x_2\right)^2\) <=> \(x_0=x_2\)( x₀,x₂>0)

Vậy x₀ là một nghiệm của pt \(x^4-16x^2+32=0\)

Lời giải:
Đặt \(\sqrt[3]{-q+\sqrt{q^2+p^3}}=a; \sqrt[3]{-q-\sqrt{q^2+p^3}}=b\) thì $x_0=a+b$

Khi đó:

\(a^3+b^3=-2q\)

\(ab=\sqrt[3]{(-q+\sqrt{q^2+p^3})(-q-\sqrt{q^2+p^3})}=\sqrt[3]{(-q)^2-(q^2+p^3)}=\sqrt[3]{-p^3}=-p\)

Ta có:

\((a+b)^3=a^3+b^3+3ab(a+b)\)

\(\Leftrightarrow x_0^3=-2q+3.(-p)x_0\)

\(\Leftrightarrow x_0^3+3px_0+2q=0\)

Do đó $x_0$ là nghiệm của PT \(x^3+3px+2q=0\)

Ta có đpcm.

x₀+y₀+\(\frac{\sqrt{3}}{2}\)=\(\frac{1}{2}\)

Nhân pt (1) với căn3 + 1 để ra 2y

\(sin9x-\sqrt{3}cos9x=sin7x-\sqrt{3}cos7x\)

\(\Leftrightarrow\frac{1}{2}sin9x-\frac{\sqrt{3}}{2}cos9x=\frac{1}{2}sin7x-\frac{\sqrt{3}}{2}cos7x\)

\(\Leftrightarrow sin\left(9x-\frac{\pi}{3}\right)=sin\left(7x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-\frac{\pi}{3}=7x-\frac{\pi}{3}+k2\pi\\9x-\frac{\pi}{3}=\frac{4\pi}{3}-7x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{5\pi}{48}+\frac{k\pi}{8}\end{matrix}\right.\)

\(\Rightarrow\) Nghiệm âm lớn nhất \(x=-\frac{\pi}{48}\)

\(y'=\dfrac{1}{2\sqrt{x-1}}+\dfrac{1}{\sqrt{2x+1}}\)

\(\Rightarrow y'\left(3\right)=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{\sqrt{7}}\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\Rightarrow a+b=\dfrac{3}{2}\)