Question

Giúp mình bài 2 với bài 3

Answer 1

Bài 3:

a. \(\sqrt{9\left(x+3\right)}-\dfrac{1}{4}\sqrt{16\left(x+3\right)}+\sqrt{x+3}=6\)

\(\Leftrightarrow3\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+3}=6\)

\(3\sqrt{x+3}=6\)

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

\(\left\{{}\begin{matrix}2>0\left(ld\right)\\x+3=4\end{matrix}\right.\Leftrightarrow x=1\)

b. \(\sqrt{2x-1}=3\)

\(\left\{{}\begin{matrix}3>0\left(ld\right)\\2x-1=9\end{matrix}\right.\Leftrightarrow x=5\)

Answer 2

Bài 2:

\(a,B=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+2\right]:\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-1\right]=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\\ b,B< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-1< 0\\ \Leftrightarrow\dfrac{\sqrt{x}+2-\sqrt{x}+1}{\sqrt{x}-1}< 0\\ \Leftrightarrow\dfrac{3}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)

Bài 3:

\(a,ĐK:x\ge-3\\ PT\Leftrightarrow3\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+3}=6\\ \Leftrightarrow3\sqrt{x+3}=6\\ \Leftrightarrow\sqrt{x+3}=2\\ \Leftrightarrow x+3=4\\ \Leftrightarrow x=1\left(tm\right)\\ b,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x-1}=3\Leftrightarrow2x-1=9\\ \Leftrightarrow x=5\left(tm\right)\)