a) ĐK: \(x^2+7x+7\ge0\)

Đặt \(a=\sqrt{x^2+7x+7}\)  \(\left(a\ge0\right)\)

PT \(\Rightarrow3a^2-3+2a=2\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2+7x+7=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)  (Thỏa mãn) 

Vậy ...

b) ĐK: \(x^2-6x+6\ge0\)

Đặt \(a=\sqrt{x^2-6x+6}\)  \(\left(a\ge0\right)\)

PT \(\Rightarrow a^2+3=4a\) \(\Leftrightarrow\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)  (Thỏa mãn)

+) Với \(a=3\) \(\Rightarrow x^2-6x+6=9\) \(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{3}\\x=3-2\sqrt{3}\end{matrix}\right.\)  (Thỏa mãn)

+) Với \(a=1\) \(\Rightarrow x^2-6x+6=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)  (Thỏa mãn)

  Vậy ...

c)C1: Áp dụng bđt AM-GM \(\Rightarrow VT\ge2>\dfrac{7}{4}\)

=> Dấu = ko xảy ra hay pt vô nghiệm

C2: Đk:\(x>0\)

Đặt \(a=\sqrt{\dfrac{x^2+x+1}{x}}\left(a>0\right)\) \(\Rightarrow\dfrac{1}{a}=\sqrt{\dfrac{x}{x^2+x+1}}\)

Pttt: \(a+\dfrac{1}{a}=\dfrac{7}{4}\Leftrightarrow4a^2-7a+4=0\) 

\(\Delta =-15<0 \) => Pt vô nghiệm

Vậy...

d) Đk: \(x\le-8;x\ge0\)

Đặt \(t=\sqrt{x\left(8+x\right)}\left(t\ge0\right)\)

Pttt: \(t^2-3=2t\Leftrightarrow t^2-2t-3=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-1\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x\left(8+x\right)}=3\Leftrightarrow x^2+8x-9=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\) (tm)

Vậy...

Ta có \(x^3=4+2\sqrt{2}+4-2\sqrt{2}+3\sqrt[3]{\left(4+2\sqrt{2}\right)\left(4-2\sqrt{2}\right)}x\)

=> \(x^3=8+6x\Rightarrow x^3-6x-10=-2\Rightarrow\left(x^3-6x-10\right)=-2^{2019}\)

^_^

kết quả lớn quá

1, \(x^3=\left(7+\sqrt{\frac{49}{8}}\right)+\left(7-\sqrt{\frac{49}{8}}\right)+3x\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}\)

\(=14+3x\cdot\frac{7}{2}=14+\frac{21x}{2}\)

\(\Leftrightarrow x^3-\frac{21}{2}x-14=0\)

Ta có: \(f\left(x\right)=\left(2x^3-21-29\right)^{2019}=\left[2\left(x^3-\frac{21}{2}x-14\right)-1\right]^{2019}=\left(-1\right)^{2019}=-1\)

2, ta có: \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\) (bạn tự cm)

Áp dụng công thức trên ta được n=2016

3, \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\frac{\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}\)

\(=\frac{\sqrt[3]{\left(\sqrt{5}-2\right)^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{\sqrt{5}+3-\sqrt{5}}=\frac{5-4}{3}=\frac{1}{3}\)

Thay x=1/3 vào A ta được;

\(A=3x^3+8x^2+2=3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2=3\)

Bài 4

ÁP DỤNG BĐT CAUCHY 

là ra

\(\frac{1}{\sqrt{1^3+2^3}}+\frac{1}{\sqrt{1^3+2^3+3^3}}+...+\frac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\frac{2015}{2017}\) (1) 

Cần CM: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\) quy nạp nhé bn, trên mạng có nhìu 

(1) \(\Leftrightarrow\)\(\frac{1}{\sqrt{\left(1+2\right)^2}}+\frac{1}{\sqrt{\left(1+2+3\right)^2}}+...+\frac{1}{\sqrt{\left(1+2+3+...+n\right)^2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+n}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+...+\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(n=2016\)

ta có : \(x=\sqrt{\sqrt{3}-\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{\sqrt{3}-\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

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

\(\Rightarrow A=\left(21x^2+6x\right)^{2018}=\left(21\left(1\right)^2+6.1\right)^{2018}=27^{2018}\)

a,ĐK: x≥4

Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}=4\)

      \(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)

b, ĐK: x≥2

Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)

      \(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)

      \(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)

Đề bài không chính xác rồi em

Muốn khử được căn ba thì trong biểu thức \(\left(2x^2-6x+2008\right)^{...}\) phải có bậc 3, mà ở đây chỉ có bậc 2

\(x=\sqrt[3]{3+2\sqrt[]{2}}+\sqrt[3]{3-2\sqrt[]{2}}\)

\(x^3=6+3\left(\sqrt[3]{3+2\sqrt[]{2}}+\sqrt[3]{3-2\sqrt[]{2}}\right)\sqrt[3]{\left(3+2\sqrt[]{2}\right)\left(3-2\sqrt[]{2}\right)}\)

\(x^3=6+3x\)

\(x^3-3x=6\)

\(P=\left[2\left(x^3-3x\right)+2008\right]^{2020}=\left(2.6+2008\right)^{2020}=2020^{2020}\)