Question

Пожалуйста, помогите с решением.

x^2+(10-x)^2=58

4x(10-x)=x^2

x^2+21=10x

x/3×x/4=x+24

Answer 1

x²+(10-x)²=58,
x²+100-20x+x²-58=0,
2x²-20x+42=0,
x²-10x+21=0,
D=100-84=16,
x1=7,
x2=3.

4x(10-x)=x²,
x²-4x(10-x)=0,
x²-40x+4x²=0,
5x²-40x=0,
x(5x-40)=0,
x=0,
5x=-40,
x=8.

x²+21=10x,
x²-10x+21=0,
D=100-84=16
x1=7,
x2=3.

x/3*x/4=x+24,
x²/12=x+24,
x²-12x-288=0,
D=144+1152=1296=36²,
x1=(12+36)2=24,
x2=-12.