Question

tìm x:

\sqrt(8x-4)-2\sqrt(18x-9)+2\sqrt(32x-16)=12

Answer 1

`\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12`      `ĐK: x >= 1/2`

`<=>2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12`

`<=>4\sqrt{2x-1}=12`

`<=>\sqrt{2x-1}=3`

`<=>2x-1=9`

`<=>x=5` (t/m)

Vậy `S={5}`.

Answer 2

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

=>\(4\sqrt{2x-1}=12\)

=>\(\sqrt{2x-1}=3\)

=>2x-1=9

=>2x=10

=>x=5

Answer 3

\(\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

\(\Leftrightarrow2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12\)

\(\Leftrightarrow\left(2-6+8\right)\sqrt{2x-1}=12\)

\(\Leftrightarrow4\sqrt{2x-1}=12\)

\(\Leftrightarrow\sqrt{2x-1}=12:4\)

\(\Leftrightarrow\sqrt{2x-1}=3\)

\(\Leftrightarrow2x-1=9\)

\(\Leftrightarrow2x=9+1\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\left(tm\right)\)

Vậy \(x=5\)