Câu hỏi của have a happy day .(^-^).

Đặt \(x=\dfrac{abc}{d}+\dfrac{bcd}{a}=\dfrac{bc\left(a^2+d^2\right)}{ad}\)\(y=\dfrac{cda}{b}+\dfrac{dab}{c}=\dfrac{ad}{bc}\left(b^2+c^2\right)\)x+y nguyên=>(x+y)^2=x^2+y^2+2xy thuộc Zmà \(x\cdot y=\left(a^2+d^2\right)\left(b^2+c^2\right)\)xy nguyên=>x^2+y^2 nguyên=>x,y nguyen=>abc/d+bcd/a nguyênmà abc/d*bcd/a=(bc)^2 nguyênnên abc/d nguyên=>abc chia hết cho dCâu trả lời: Đặt \(x=\dfrac{abc}{d}+\dfrac{bcd}{a}=\dfrac{bc\left(a^2+d^2\right)}{ad}\)\(y=\dfrac{cda}{b}+\dfrac{dab}{c}=\dfrac{ad}{bc}\left(b^2+c^2\right)\)x+y nguyên=>(x+y)^2=x^2+y^2+2xy thuộc Zmà \(x\cdot y=\left(a^2+d^2\right)\left(b^2+c^2\right)\)xy nguyên=>x^2+y^2 nguyên=>x,y nguyen=>abc/d+bcd/a nguyênmà abc/d*bcd/a=(bc)^2 nguyênnên abc/d nguyên=>abc chia hết cho d