Résoudre les équations: (3x+2 )/5-5x-7= (-8x+3)/25 x(4x-7)(9-2x)(8-2/7x)=0 (x+7)²=144

Bonsoir,

[tex]1)\ \dfrac{3x+2}{5}-5x-7= \dfrac{-8x+3}{25}\\\\\dfrac{5(3x+2)}{25}-\dfrac{125x}{25}-\dfrac{175}{25}= \dfrac{-8x+3}{25}\\\\\dfrac{5(3x+2)-125x-175}{25}= \dfrac{-8x+3}{25}\\\\5(3x+2)-125x-175=-8x+3\\\\15x+10-125x-175=-8x+3\\\\15x-125x+8x=3-10+175\\\\-102x=168[/tex]

[tex]x=-\dfrac{168}{102}\\\\\boxed{x=-\dfrac{28}{17}\approx1,647}[/tex]

[tex]2)\ x(4x-7)(9-2x)(8-\dfrac{2}{7}x)=0\\\\x=0\ \ ou\ \ 4x-7=0\ \ ou\ \ 9-2x=0\ \ ou\ \ 8-\dfrac{2}{7}x=0\\\\x=0\ \ ou\ \ 4x=7\ \ ou\ \ -2x=-9\ \ ou\ \ -\dfrac{2}{7}x=-8\\\\x=0\ \ ou\ \ x=\dfrac{7}{4}\ \ ou\ \ x=\dfrac{-9}{-2}\ \ ou\ \ x=-8\times(-\dfrac{7}{2})\\\\x=0\ \ ou\ \ x=\dfrac{7}{4}\ \ ou\ \ x=\dfrac{9}{2}\ \ ou\ \ x=\dfrac{56}{2}\\\\x=0\ \ ou\ \ x=1,75\ \ ou\ \ x=4,5\ \ ou\ \ x=28\\\\\boxed{S=\{0\ ;\ 1,75\ ;\ 4,5\ ;\ 28\}}[/tex]

[tex]3)\ (x+7)^2=144\\\\(x+7)^2-144=0\\\\(x+7)^2-12^2=0\\\\ \ [(x+7)+12][(x+7)-12]=0\\\\(x+7+12)(x+7-12)=0\\\\(x+19)(x-5)=0\\\\x+19=0\ \ ou\ \ x-5=0[/tex]

[tex]\\\\x=-19\ \ ou\ \ x=5\\\\\boxed{S=\{-19\ ;\ 5\}}[/tex]