urgent merci d avance

Bonjour Mademoiselle123

[tex]1)\ (2x+1)^2=25\\(2x+1)^2-25=0\\(2x+1)^2-5^2=0\\ \ [(2x+1)-5][(2x+1)+5]=0\\(2x+1-5)(2x+1+5)=0\\(2x-4)(2x+6)=0\\2x-4=0\ \ ou\ \ 2x+6=0\\2x=4\ \ ou\ \ 2x=-6\\\boxed{x=2\ \ ou\ \ x=-3}[/tex]

[tex]2)\ 3x^2-4x=0\\x(3x-4)=0\\x=0\ \ ou\ \ 3x-4=00\\x=0\ \ ou\ \ 3x=4\\\boxed{x=0\ \ ou\ \ x=\dfrac{4}{3}}[/tex]

[tex]3)\ (2x-\dfrac{3}{4})(3x^2-4x)=0\\\\2x-\dfrac{3}{4}=0\ \ ou\ \ 3x^2-4x=0\\\\2x=\dfrac{3}{4}\ \ ou\ \ x(3x-4)=0\\\\x=\dfrac{3}{8}\ \ ou\ \ x=0\ \ ou\ \ 3x-4=0\\\\x=\dfrac{3}{8}\ \ ou\ \ x=0\ \ ou\ \ 3x=4\\\\\boxed{x=\dfrac{3}{8}\ \ ou\ \ x=0\ \ ou\ \ x=\dfrac{4}{3}}[/tex]

[tex]4)\ 4x^2-12x+9=0\\(2x-3)^2=0\\2x-3=0\\2x=3\\\boxed{x=\dfrac{3}{2}}[/tex]

[tex]5)\ 4-(x-3)^2=0\\2^2-(x-3)^2=0\\\ [2-(x-3)][2+(x-3)]=0\\(2-x+3)(2+x-3)=0\\(-x+5)(x-1)=0\\-x+5=0\ \ ou\ \ x-1=0\\-x=-5\ \ ou\ \ x=1\\\boxed{x=5\ \ ou\ \ x=1}[/tex]

[tex]6)\ 9x^2+6x+1=0\\(3x+1)^2=0\\3x+1=0\\3x=-1\\\boxed{x=-\dfrac{1}{3}}[/tex]

[tex]7)\ (\dfrac{25}{4}x^2-1)(x+2)=0\\\\(\dfrac{25}{4}x^2-1^2)(x+2)=0\\\\(\dfrac{5}{2}x-1)(\dfrac{5}{2}x+1)(x+2)=0\\\\\dfrac{5}{2}x-1=0\ \ ou\ \ \dfrac{5}{2}x+1=0\ \ ou\ \ x+2=0\\\\\dfrac{5}{2}x=1\ \ ou\ \ \dfrac{5}{2}x=-1\ \ ou\ \ x=-2\\\\\boxed{x=\dfrac{2}{5}\ \ ou\ \ x=-\dfrac{2}{5}\ \ ou\ \ x=-2}[/tex]

[tex]8)\ 5x-15x^2=0\\5x(1-3x)=0\\5x=0\ \ ou\ \ 1-3x=0\\x=\dfrac{0}{5}\ \ ou\ \ -3x=-1\\x=0\ \ ou\ \ x=\dfrac{-1}{-3}\\\boxed{x=0\ \ ou\ \ x=\dfrac{1}{3}}[/tex]

[tex]9)\ 3x^2-4=0\\(\sqrt{3}x)^2-2^2=0\\(\sqrt{3}x-2)(\sqrt{3}x+2)=0\\\sqrt{3}x-2=0\ \ ou\ \ \sqrt{3}x+2=0\\\sqrt{3}x=2\ \ ou\ \ \sqrt{3}x=-2\\x=\dfrac{2}{\sqrt{3}}\ \ ou\ \ x=-\dfrac{2}{\sqrt{3}}\\\\\boxed{x=\dfrac{2\sqrt{3}}{3}\ \ ou\ \ x=-\dfrac{2\sqrt{3}}{3}}[/tex]

[tex]10)\ Enonc\acute{e}\ incomplet...\\\\(2-3x)(...)=0\\2-3x=0\ \ ou\ \ ...=0\\-3x=-2\ \ ou\ \ ...\\x=\dfrac{-2}{-3}\ \ ou\ \ ...\\\\\boxed{x=\dfrac{-2}{-3}\ \ ou\ \ x=...}[/tex]