ĐKXĐ: x>=0; x<>1

PT =>\(\dfrac{\left(\sqrt{x}+3\right)\left(-2x+6\right)}{\left(\sqrt{x}-1\right)^2}=0\)

=>6-2x=0

=>x=3

\(\Leftrightarrow2x^3-3x^2+6x+2x^2-3x+6=0\)

\(\Leftrightarrow x\left(2x^2-3x+6\right)+2x^2-3x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2-3x+6=0\left(vn\right)\end{matrix}\right.\)

Câu a :

\(x^2-2x-3=0\)

\(\Leftrightarrow x^2-x+3x-3=0\)

\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\x+3=0\Rightarrow x=-3\end{matrix}\right.\)

Câu b :

\(2x^2+3=-5x\)

\(\Leftrightarrow2x^2+3+5x=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\2x+3=0\Rightarrow x=-\dfrac{3}{2}\end{matrix}\right.\)

Mấy câu sau khó quá ko bt làm :)

\(x^3+2x-3=0\)

\(\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(3x-3\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+3\right)=0\)

Ta có: \(x^2+x+3=\left[x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+3-\left(\dfrac{1}{2}\right)^2\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\) (1)

Mà \(\left(x-1\right)\left(x^2+x+3\right)=0\) từ (1) \(\Rightarrow x-1=0\Leftrightarrow x=1\)

Vậy x = 1

\(x^3-x+3x-3=0\)

\(\Leftrightarrow x\left(x^2-1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)+3\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vl\right)\end{matrix}\right.\)

vậy \(S=\left\{1\right\}\)

a) (2x - 1) x - x (2x + 3) = 7

<=> x (2x - 1 - 2x - 3) = 7

<=> -4x = 7

<=> x = \(-\dfrac{7}{4}\)

b) 3 (2x - 1) - 5 (x - 3) + 6 (3x - 4) = 24

<=> 6x - 3 - 5x + 15 + 18x - 24 = 24

<=> 19x - 12 = 24

<=> 19x = 36

<=> x = \(\dfrac{36}{19}\)

a) 

b)

c)

e)

Những câu còn lại mk hổng bt làm đâu