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

\(=x\left(\sqrt{x}-1\right)-2\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)\)

\(=\left(\sqrt{x}-1\right)\left(x-2\sqrt{x}+2\right)\)

\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)

\(=2x^2-2x\sqrt{x+3}-x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)

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

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

\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)

\(=2x^2-x\sqrt{x+3}-2x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)

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

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

\(=x+2\sqrt{xy}+y-9\)

\(=\left(\sqrt{x}+\sqrt{y}\right)^2-3^2\)

\(=\left(\sqrt{x}+\sqrt{y}-3\right)\left(\sqrt{x}+\sqrt{y}+3\right)\)

Cảm ơn bạn nha

Mình làm được rồi

Câu b đề sai nha bạn.

a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)

b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)

c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

Lời giải:

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

$=x(\sqrt{x}-1)+2\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)$

$=(\sqrt{x}-1)(x+2\sqrt{x}+2)$

Đến đây là PT tích r còn gì, \(x\in\left\{5;-10;-\sqrt{3}\right\}\)

:v ez mà :) Sửa đề : tìm x 

\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)

TH1 : \(2x-10=0\Leftrightarrow x=5\)

TH2 : \(x+10=0\Leftrightarrow x=-10\)

TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )

Vậy x = 5 ; x = -10