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

\(2\sqrt{2x}\)

3√2x−√18x+12√32x

=\(3\sqrt{2x}-3\sqrt{2x}+48\sqrt{2x}\)

=\(48\sqrt{2x}\)

\(=3\sqrt{2x}-3\sqrt{2x}+\dfrac{1}{2}\cdot4\sqrt{2x}\)

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

\(3\sqrt{2x}-\sqrt{18x}+\dfrac{1}{2}\sqrt{32x}\)

\(=3\sqrt{2x}-3\sqrt{2x}+\dfrac{1}{2}\cdot4\sqrt{2x}\)

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

\(M=\dfrac{3}{2}\cdot4\sqrt{2x}-\dfrac{1}{3}\cdot3\sqrt{2x}+\dfrac{2}{5}\cdot5\sqrt{2x}-4\sqrt{2x}=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)

\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)

ĐKXĐ: x>=-1/2

\(2\sqrt{32x+16}-3\sqrt{18x+9}=\sqrt{8x+4}-6\)

=>\(2\cdot4\sqrt{2x+1}-3\cdot3\sqrt{2x+1}-2\sqrt{2x+1}=-6\)

=>\(8\sqrt{2x+1}-9\sqrt{2x+1}-2\sqrt{2x+1}=-6\)

=>\(-3\sqrt{2x+1}=-6\)

=>\(\sqrt{2x+1}=2\)

=>2x+1=4

=>2x=3

=>\(x=\dfrac{3}{2}\left(nhận\right)\)

\(ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow3\sqrt{2x-3}-2\sqrt{2x-3}+6\sqrt{2x-3}=1\\ \Leftrightarrow7\sqrt{2x-3}=1\\ \Leftrightarrow\sqrt{2x-3}=\dfrac{1}{7}\\ \Leftrightarrow2x-3=\dfrac{1}{49}\Leftrightarrow x=\dfrac{74}{49}\left(tm\right)\)

h: \(\sqrt{18x}+\sqrt{32x}-14=0\)

\(\Leftrightarrow7\sqrt{2x}=14\)

hay x=2

`\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}`.

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

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

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

=>2x-1=9

=>2x=10

=>x=5

\(\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12\) (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 \(x=5\)