Question

Задание в приложении , 50 баллов.

Answer 1

(x₁-x₀)(x₂`-x₀`)=(x₂-x₀)(x₁-x₀`)
x₁x₂`-x₁x₀`-x₀x₂`+x₀x₀`=x₁x₂-x₂x₀`-x₁x₀+x₀x₀`
x₁x₂`-x₁x₀`-x₀x₂`-x₁x₂+x₂x₀`+x₁x₀=0
x₂x₀`-x₁x₀`=-x₁x₂`+x₀x₂`+x₁x₂-x₁x₀
x₀`(x₂-x₁)=-x₁x₂`+x₀x₂`+x₁x₂-x₁x₀
x₀`=(-x₁x₂`+x₀x₂`+x₁x₂-x₁x₀)/(x₂-x₁)

Answer 2

$\frac{ x_{1} - x_{0} }{ x_{2}- x_{0} } = \frac{ x_{1} -x _{0} ^{'} }{ x_{2} ^{'} - x_{0} ^{'} } \\ \\ ( x_{1} - x_{0} )( x_{2} ^{'} - x_{0} ^{'} )=( x_{2} - x_{0} )( x_{1} - x_{0} ^{'} ) \\ \\ x_{1} x_{2} ^{'} - x_{1} x_{0} ^{'} - x_{2} ^{'} x_{0} + x_{0} x_{0} ^{'} = x_{1} x_{2} - x_{2} x_{0} ^{'} - x_{1} x_{0} + x_{0} x_{0} ^{'}$


$x_{1} x_{2} ^{'} - x_{1} x_{0} ^{'} - x_{2} ^{'} x_{0} + x_{0} x_{0} ^{'} - x_{1} x_{2}+ x_{2} x_{0} ^{'} + x_{1} x_{0} - x_{0} x_{0} ^{'} =0 \\ \\ - x_{1} x_{0} ^{'} + x_{0} x_{0} ^{'} - x_{0} x_{0} + x_{2} x_{0} ^{'} =- x_{1} x_{2} ^{'} + x_{2} ^{'} x_{0} + x_{1} x_{2} - x_{1} x_{0} \\ \\ - x_{1} x_{0} ^{'} + x_{2} x_{0} ^{'} =- x_{1} x_{2} ^{'} + x_{2} ^{'} x_{0} + x_{1} x_{2} - x_{1} x_{0}$ 

$\\ \\ x_{0} ^{'} (- x_{1} + x_{2} )= - x_{1} x_{2} ^{'} + x_{2} ^{'} x_{0} + x_{1} x_{2} - x_{1} x_{0} \\ \\ x_{0} ^{'} = \frac{- x_{1} x_{2} ^{'} + x_{2} ^{'} x_{0} + x_{1} x_{2} - x_{1} x_{0}}{- x_{1} + x_{2} } \\ \\ x_{0} ^{'} = \frac{-( x_{1} x_{2} ^{'} - x_{2} ^{'} x_{0} - x_{1} x_{2} + x_{1} x_{0})}{- (x_{1} - x_{2}) } \\ \\ x_{0} ^{'} = \frac{ x_{1} x_{2} ^{'} - x_{2} ^{'} x_{0} - x_{1} x_{2} + x_{1} x_{0}}{x_{1} - x_{2} }$