\(\left(\sqrt{1+\sqrt{1+\sqrt{1}}}\right)^4\)

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

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

\(=\left[\left(\sqrt{1+\sqrt{2}}\right)^2\right]^2\)

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

\(=1^2+2\cdot\sqrt{2}\cdot1+\left(\sqrt{2}\right)^2\)

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

\(=3+2\sqrt{2}\)

\(a^2+2a\left(1+x\right)-8x-8=0\\ \Leftrightarrow a^2+2a\left(1+x\right)-8\left(x+1\right)=0\\ \Leftrightarrow\left(a^2+2a\right)-8\left(x+1\right)=0\\ \Leftrightarrow a\left(a+2\right)-8\left(x+1\right)=0\\ \Leftrightarrow\left(a-8\right)\left(a+2\right)\left(x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}a-8=0\\a+2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=8\\a=-2\\x=-1\end{matrix}\right.\)

Chỗ a(a+2) - 8(x+1) sao đặt nhân tử chung được vậy 

Ta có:

\(\sqrt{2}\approx1,414214,...\) 

\(\sqrt{3}\approx1,732051...\)

Nên số hữu tỉ giữa hai số là: \(1,5=\dfrac{3}{2}\)

Mà: \(\sqrt{2}< \sqrt{2,5}< \sqrt{3}\)

Nên số vô tỉ giữa hai số là: \(\sqrt{2,5}\approx1,58...\)

Bài 5:

Ta có: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

\(\Rightarrow\sqrt{A}=\sqrt{\dfrac{\sqrt{x}-2}{\sqrt{x}+1}}\)

Mà: \(\sqrt{A}< \dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{\dfrac{\sqrt{x}-2}{\sqrt{x}+1}}< \dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< \dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{9\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)}{9\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{9\sqrt{x}-18-\sqrt{x}-1}{9\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{8\sqrt{x}-19}{9\left(\sqrt{x}+1\right)}< 0\Leftrightarrow8\sqrt{x}-19< 9\)

\(\Leftrightarrow8\sqrt{x}< 19\)

\(\Leftrightarrow\sqrt{x}< \dfrac{19}{8}\)

\(\Leftrightarrow x< \dfrac{361}{64}\)

Kết hợp với đk:

\(0\le x< \dfrac{361}{64}\)

mik cần bài 6 cơ ạ;-;