Question

Скиньте кто может решение пожалуйста

Answer 1

$x^2-7x+3=0\\ D=(-7)^2 -4 \cdot 1 \cdot 3=49-12=37\\ x_1= \frac{7+ \sqrt{37} }{2} \\ x_2= \frac{7- \sqrt{37} }{2} \\ x_1^2= (\frac{7+ \sqrt{37} }{2})^2= \frac{49+14 \sqrt{37}+37 }{4}= \frac{86+14 \sqrt{37} }{4}= \frac{43+7 \sqrt{37} }{2}\\ x_2^2= (\frac{7- \sqrt{37} }{2})^2= \frac{49-14 \sqrt{37}+37 }{4}= \frac{86-14 \sqrt{37} }{4}= \frac{43+7 \sqrt{37} }{2}$
$\frac{1}{x_1^2} + \frac{1}{x_2^2} = \frac{2}{43+7 \sqrt{37} } +\frac{2}{43-7 \sqrt{37} }= \frac{2(43-7 \sqrt{37} )+2(43+7 \sqrt{37} )}{(43+7 \sqrt{37} )(43-7 \sqrt{37}) } =\\ = \frac{86-14 \sqrt{37}+86+14 \sqrt{37} }{43^2-(7 \sqrt{37} )^2 }= \frac{172}{1849-1813}= \frac{172}{36}= \frac{43}{9}$

Ответ:4)

Answer 2

$x^{2} - 7x+3=0$
D = 49 - 4*3 = 49 - 12 = 37
$x_{1}= \frac{7- \sqrt{37} }{2}$
$x_{2}= \frac{7+ \sqrt{37} }{2}$
$\frac{1}{ x_{1} ^{2} } + \frac{1}{ x_{2} ^{2} }= \frac{1}{ (\frac{7- \sqrt{37} }{2}) ^{2} }+ \frac{1}{ ( \frac{7+ \sqrt{37} }{2} )^{2} }= \frac{4* (7+ \sqrt{37}) ^{2} +4* (7- \sqrt{37})^{2} }{ (7- \sqrt{37}) ^{2} * (7+ \sqrt{37}) ^{2} }= \frac{688}{ (49-37)^{2} }= \frac{43}{9}$
Ответ: 4.