Question

Mn giải giúp e câu c bài 2 đấy thôi ạ

Answer 1

c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=5\)

\(\Leftrightarrow\sqrt{x-4}+2=5\)

\(\Leftrightarrow\sqrt{x-4}=3\)

\(\Leftrightarrow x-4=9\)

hay x=13

Answer 2

c: Ta có: 

hay x=13

 

Answer 3

\(a,ĐK:x\ge-1\\ PT\Leftrightarrow2x^2+2x-4=x^2+2x+1\\ \Leftrightarrow x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\left(tm\right)\\x=-\sqrt{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\sqrt{5}\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-2\right|=2x-5\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-5\left(x\ge2\right)\\x-2=5-2x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\dfrac{7}{3}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=3\)

\(c,ĐK:x\ge4\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=5\\ \Leftrightarrow\sqrt{x-4}=3\Leftrightarrow x-4=9\Leftrightarrow x=5\left(tm\right)\)