Question

2×(х-5)-12=3 (1-х)
х-10=5 (2х+1)+57
3×(х-5)+10×(х-8)=30
2×(х+1)+3×(х-2)=24

Answer 1

$\ 2x(x-5)-12=3(1-x) \ 2x^2-10x-12=3-3x \ 2x^2-10x-12-3+3x=0 \ 2x^2-7x-9=0 \ D=(-7)^2-4*2*(-9)=49+72=121 \ sqrt{121}=11 \ x1= frac{7+11}{2*2 }= frac{20}{4}= 5 \ x2= frac{7-11}{4}=- frac{4}{4}=-1$

$\ x-10=5(2x+1)+57 \ x-10=10x+5+57 \ x-10x=62+10 \ -9x=72 \ x= frac{72}{-9} \ x=-8$

Answer 2

Спасибо большое