Question

[vy phan ngọc vy]

tìm nghiệm nguyên của PT:x²+(x+1)²=y⁴+(y+1)⁴

nhanh nhes

Answer 1

[TRẦN ĐỨC VINH]

x\(=\)y\(=\)0      va   băng âm một 

Answer 2

[Nguyễn Ngọc Minh]

\(x^2+\left(x+1\right)^2=y^4+\left(y+1\right)^4\)

\(\Leftrightarrow x^2-y^4=\left(y+1\right)^4-\left(x+1\right)^2\)

\(\Leftrightarrow x^2-\left(y^2\right)^2=\left(\left(y+1\right)^2\right)^2-\left(x+1\right)^2\)

\(\Leftrightarrow\left(x-y^2\right)\left(x+y^2\right)=\left(\left(y+1\right)^2+x+1\right)\left(\left(y+1\right)^2-x-1\right)\)

\(\Leftrightarrow\left(x-y^2\right)\left(x+y^2\right)=\left(y^2+2y+1+x+1\right)\left(y^2+2y+1-x-1\right)\)

\(\Leftrightarrow\left(x-y^2\right)\left(x+y^2\right)=\left(y^2+2y+x+2\right)\left(y^2+2y-x\right)\)

\(\Leftrightarrow\frac{\left(x-y^2\right)}{\left(y^2+2y-x\right)}=\frac{\left(y^2+2y+x+2\right)}{\left(x+y^2\right)}\)

Đến đây dễ dàng tính rồi nhé -_<