Question

Tìm x; y biết: x³+x²+x+1=y^3

HELP, HELP!!!!!

Answer 1

help me please

Answer 2

Có: x²+x+1\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\) với mọi x

=>x³+x²+x+1>x³

=>y³>x³  (1)

Lại có (x+2)³-(x³+x²+x+1)

=x³+8+6x²+12x-x³-x²-x-1=5x²+11x+7=\(5\left(x^2+\frac{11}{5}x+\frac{7}{5}\right)=5\left(x^2+2.x.\frac{11}{10}+\frac{121}{100}+\frac{19}{100}\right)=5\left(x+\frac{11}{10}\right)^2+\frac{19}{20}\ge\frac{19}{20}>0\) với mọi x

=>(x+2)³ \(\ge\) x³+x²+x+1  (2)

Từ (1),(2)

=>x³<y³<(x+2)³

=>y³=(x+1)³ => x³+x²+x+1=(x+1)³

=>x²(x+1)+(x+1)-(x+1)³=0

=>(x²+1)(x+1)-(x+1)³=0

=>(x+1)x=0=>x=0 hoặc x=-1

+x=0 thì y=1

+x=-1 thì y=0

Vậy (x;y)=...............