Question

[Sakura]

a)Tính: \(\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\sqrt{2}+2}\)

b)Giải phương trình sau :\(\sqrt{x^2-12x+36}=81\)

c)Giải phương trình sau : \(\sqrt{4x^2-12x+9}=7\)

Answer 1

[Nguyễn Việt Lâm]

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

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

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

b/ \(\Leftrightarrow x^2-12x+36=6561\)

\(\Leftrightarrow x^2-12x-6525=0\)

\(\Leftrightarrow\left(x-87\right)\left(x+75\right)=0\Rightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right.\)

c/ \(\Leftrightarrow4x^2-12x+9=49\)

\(\Leftrightarrow4x^2-12x-40=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Hai câu b; c đều có thể giải bằng cách sử dụng hằng đẳng thức, nhưng cần phá trị tuyệt đối tốn thời gian, tốt nhất là bình phương cho lẹ

Answer 2

[svtkvtm]

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

\(Đat:A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\Rightarrow A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}+2=2\left(\sqrt{2}+1\right)\Rightarrow A=\sqrt{2\sqrt{2}+2}\left(vì:\sqrt{\sqrt{2}-1};\sqrt{\sqrt{2}+1}>0\right)\) \(\Rightarrow\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\sqrt{2}+2}=\sqrt{2\sqrt{2}+2}-\sqrt{2\sqrt{2}+2}=0\)

\(b,\sqrt{x^2-12x+36}=\sqrt{\left(x-6\right)^2}=\left|x-6\right|=81\Leftrightarrow\left[{}\begin{matrix}x-6=81\\x-6=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right..Vậy:x\in\left\{87;-75\right\}\)

\(c,\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=7\Leftrightarrow\left|2x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}2x-3=-7\\2x-3=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right..Vậy:x\in\left\{-2;5\right\}\)