Question

144z^2-1=0; (4t-3)^2-25t^2; (a+1)^2-(2a+3)^2=0. все три примера плз

Answer 1

$144z^2-1=0\\ 144z^2=1\\ z^2= \frac{1}{144}\\\\ z_1= \frac{1}{12}\\\\ z_1= - \frac{1}{12}$

ответ: $z_1=\frac{1}{12}; z_2=- \frac{1}{12}$

$(4t-3)^2-25t^2=0\\ 16t^2-24t-19-25t^2=0\\ -9t^2-24t+9=0\ \ |:(-3)\\ 3t^2+8t-3=0\\ D=64+36=100; \ \sqrt{D}=10\\\\ x_{1/2}= \frac{-8\pm10}{6}\\\\ x_1=-3\\ x_2= \frac{1}{3}$

ответ: $x_1=-3; x_2=\frac{1}{3}$

$(a+1)^2-(2a+3)^2=0\\ a^2+2a+1-(4a^2+12a+9)=0\\ -3a^2-10a-8=0\\ 3a^2+10a-8=0\\ D=100-96=4; \ \sqrt{D}=2\\\\ x_{1/2}= \frac{-10\pm2}{6}\\\\ x_1=-2\\ x_2=- \frac{4}{3}$

ответ: $x_1=-2; x_2=- \frac{4}{3}$