x³+x²-3x+1=0
x³-x²+2x²-2x-x+1=0
x²(x-1)+2x(x-1)-(x-1)=0
(x-1)(x²+2x-1)=0
1)x-1=0; x=1.
2)x²+2x-1=0
D=2²-4*1*(-1)=4+4=8=(2√2)²
x_(1)=(-2-2√2)/2=(2(-1-√2))/2=-1-√2
x_(2)=(-2+2√2)/2=(2(-1+√2))/2=-1+√2
Ответ: -1-√2; -1+√2; 1.
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x³+x²-3x+1=0
x³-x²+2x²-2x-x+1=0
x²(x-1)+2x(x-1)-(x-1)=0
(x-1)(x²+2x-1)=0
1)x-1=0; x=1.
2)x²+2x-1=0
D=2²-4*1*(-1)=4+4=8=(2√2)²
x_(1)=(-2-2√2)/2=(2(-1-√2))/2=-1-√2
x_(2)=(-2+2√2)/2=(2(-1+√2))/2=-1+√2
Ответ: -1-√2; -1+√2; 1.