Question

Answer 1

i. x-1+\(\sqrt{2x+1}\)=7

<=>

Answer 2

h)\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}=16-\sqrt{x-1}\left(đk:x\ge1\right)\)

\(\Leftrightarrow2\sqrt{x-1}=16\)

\(\Leftrightarrow\sqrt{x-1}=8\Leftrightarrow x-1=64\Leftrightarrow x=65\left(tm\right)\)

i) \(x-1+\sqrt{2x-1}=7\left(đk:8\ge x\ge\dfrac{1}{2}\right)\)

\(\Leftrightarrow8-x=\sqrt{2x-1}\)

\(\Leftrightarrow x^2-16x+64=2x-1\)

\(\Leftrightarrow\left(x-5\right)\left(x-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=13\left(l\right)\end{matrix}\right.\)