\(=\sqrt{x}\left(3\sqrt{2}-5\sqrt{8}+7\sqrt{18}\right)+28\\ =\sqrt{x}\left(3\sqrt{2}-10\sqrt{2}+21\sqrt{2}\right)+28\\ =\sqrt{x}\cdot14\sqrt{2}+28=14\sqrt{2}\left(\sqrt{x}+\sqrt{2}\right)\)

sorry mìh k bít làm

mình biết lm nhưng k biết gõ dấu căn

\(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\)28

<=> \(3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}=28\)

<=> \(14\sqrt{2x}=28\)

<=> 

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

\(< =>3\sqrt{2x}-5\sqrt{2^2.2x}+7\sqrt{3^2.2x}=28\)

\(< =>3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

\(< =>14\sqrt{2x}=28\)

\(< =>\sqrt{2x}=\dfrac{28}{14}=2=\sqrt{4}\)

\(< =>\sqrt{2x}=\sqrt{2.2}=>x=2\)

câu b bạn đã giải được chưa

Lời giải:

a. ĐKXĐ: $x\geq 0$

$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$

$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$

$\Leftrightarrow 13\sqrt{2x}=28$

$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$

$\Leftrightarrow 2x=\frac{784}{169}$

$\Leftrightarrow x=\frac{392}{169}$

b. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$

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

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$

PT $\Leftrightarrow \frac{3x-2}{x+1}=9$

$\Rightarrow 3x-2=9(x+1)$

$\Leftrightarrow x=\frac{-11}{6}$ (tm)

<=>3\(\sqrt{2x}\)-20\(\sqrt{2x}\)+21\(\sqrt{2x}\)=28

<=>4\(\sqrt{2x}\)=28

<=>\(\sqrt{2x}\)=7

<=>2x=14

<=>x=7

\(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\)

\(3\sqrt{2x}-5\sqrt{8}.\sqrt{x}+7\sqrt{18x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2}.\sqrt{x}+7\sqrt{18x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2}.\sqrt{x}+7.\sqrt{18}.\sqrt{x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2}.\sqrt{x}+7.3\sqrt{2}.\sqrt{x}=28\)

\(3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}=28\)

\(3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

\(14\sqrt{2x}=28\)

\(392x=784\)

\(x=\frac{784}{392}=2\)