Pouvez vous m'aider à résoudre ces équations svp. (3-x)^2=289 (5x-7)^2=10

Bonsoir

1) (3 - x)² = 289
(3 - x)² - 289 = 0
(3 - x)² - 17² = 0
[(3 - x) + 17][(3 - x) - 17] = 0
(3 - x + 17)(3 - x - 17) = 0
(-x + 20)(-x - 14) = 0
-x + 20 = 0   ou   -x - 14 = 0
-x = -20        ou       -x = 14
x = 20          ou       x = -14

S = {20 ; -14}


b) (5x -7)² = 10
(5x - 7)² - 10 = 0

[tex](5x-7)^2-(\sqrt{10})^2=0\\\\\ [(5x-7)+\sqrt{10}][(5x-7)-\sqrt{10}]=0\\\\\ (5x-7+\sqrt{10})(5x-7-\sqrt{10})=0\\\\\ 5x-7+\sqrt{10}=0\ \ ou\ \ 5x-7-\sqrt{10}=0\\\\\ 5x=7-\sqrt{10}\ \ ou\ \ 5x=7+\sqrt{10}\\\\\ x=\dfrac{7-\sqrt{10}}{5}\approx 0,77\ \ ou\ \ x=\dfrac{7+\sqrt{10}}{5}\approx 2[/tex]