Tìm nghiệm nguyên

H

hthtb22

Ngắn gọn hơn
$$\frac{xyzt+xy+xt+zt+1}{yzt+y+t}=x+\frac{1}{y+ \frac{1}{ t+\frac{1}{z}}}$$
Mà
$$\frac{54}{17}=3+\frac{1}{5+\frac{1}{1+\frac{1}{2}}}$$
Đồng nhất 2 biểu thức ta có:
$$x=3;y=5;t=1;z=2$$
 
H

hn3

Đề bài như này chứ nhỷ :-/ $17(xyzt+xy+xt+zt+1)=54(yzt+y+t)$

<=> $\frac{xyzt+xy+xt+zt+1}{yzt+y+t}=\frac{54}{17}$

<=> $x+\frac{zt+1}{yzt+y+t}=3+\frac{3}{17}$

==> $x=3 ; \frac{zt+1}{yzt+y+t}=\frac{3}{17}$

==> $\frac{yzt+y+t}{zt+1}=\frac{17}{3}$

<=> $y+\frac{t}{zt+1}=5+\frac{2}{3}$

==> $y=5 ; \frac{t}{zt+1}=\frac{2}{3}$

==> $t=2 ; z=1$

Vậy : $x=3 ; y=5 ; z=1 ; t=2$

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