ĐKXĐ : \(4-9x^2\ge0\Leftrightarrow\left(2-3x\right)\left(2+3x\right)\ge0\Leftrightarrow-\frac{2}{3}\le x\le\frac{2}{3}\)

\(9x^2-\left(2x-1\right)^2=\left(5x-1\right)\left(x-3\right)\)

\(\Leftrightarrow\left(3x-2x+1\right)\left(3x+2x-1\right)=\left(5x-1\right)\left(x-3\right)\)

\(\Leftrightarrow\left(x+1\right)\left(5x-1\right)-\left(5x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x+1-x+3\right)=0\Leftrightarrow x=\frac{1}{5}\)

:))))))))))))))))

:0)))

3x ( 2x - 1 ) - x ( 5x + 2 ) - x + 9 

= 6x² - 3x - 5x² - 2x - x + 9

= x² - 6x + 9

= ( x - 3 )²

b) \(\left(2x-3\right)^2\left(x-1\right)-\left(x+4\right)3x\)\(=\left(4x^2-12x+9\right)\left(x-1\right)-\left(x+4\right)3x\)

\(=4x^3-12x^2+9x-4x^2+12x-9-3x^2-12x=4x^3-19x^2+9x-9\)

c) \(\left(3x-4\right)\left(x+5\right)-\left(6x-3\right)\left(2x+1\right)\)

\(=3x^2+15x-4x-20-12x^2-6x+6x-3=-9x^2+11x-23\)

d) \(\left(x-3\right)^3+\left(x+4\right)\left(x+5\right)\)

\(=x^3-3x^2+9x-27+x^2+4x+5x+20=x^3-2x^2+18x-7\)

e) \(\left(x-4\right)\left(x+6\right)+\left(x-3\right)\left(x+2\right)\)

\(=x^2+6x-4x-24+x^2+2x-3x-6=2x^2+x-30\)

Tương tự với các phần g và h

ta có :

\(3x^2+2x-2\left(x^2-6x+5\right)+6=0\)

\(\Leftrightarrow3x^2+2x-2x^2+12x-10+6=0\)

\(\Leftrightarrow x^2+14x-4=0\Leftrightarrow\left(x^2+14x+49\right)=53\)

\(\Leftrightarrow\left(x+7\right)^2=53\Leftrightarrow x+7=\pm\sqrt{53}\Leftrightarrow x=-7\pm\sqrt{53}\)

\(x\left(3x+2\right)-2\left(x-5\right)\left(x-1\right)+6=0\)

\(\Leftrightarrow3x^2+2x-2x^2+2x+10x-10+6=0\)

\(\Leftrightarrow x^2+14x-4=0\)

\(\Leftrightarrow x^2+14x+49-53=0\)

\(\Leftrightarrow\left(x+7\right)^2-53=0\)

\(\Leftrightarrow\left(x+7-\sqrt{53}\right)\left(x+7+\sqrt{53}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+7-\sqrt{53}=0\\x+7+\sqrt{53}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{53}-7\\x=-\sqrt{53}-7\end{cases}}\)

Vậy \(S=\left\{\sqrt{53}-7;-\sqrt{53}-7\right\}\)

Đây nè

bạn kham khảo nha

HT nha