1)tìm nghiệm nguyên của phương trình: $2x^3-x^2y+3x^2+2x-y=2$
2)cho x,y,z là các số dương và x+y+z=1. Chứng minh rằng : $P=\frac{x}{1-z}+\frac{y}{1-x}+\frac{z}{1-y}<2$
3) giải hệ phương trình$\left\{ \begin{array}{l} x^3+y^3=91\\ 4x^2+3y^2=16x+9y \end{array} \right.$
còn 1 câu nữa giải nốt k nhỉ :v –  tran85295 22-12-17 03:18 PM
$pt\Leftrightarrow (x^2+1)(2x-y+3)=5$
Do x,y nguyên nên $x^2+1;2x+y-3$ nguyên
$x^2+1;2x+y+3$ là ước của 5
em chịu khó giải tiếp nhé

3
$\begin{cases}x^3+y^3=91(1) \\ 4x^2+3y^2=16x+9y(2) \end{cases}$
Lấy $(1)-3.(2)$ được $(x-4)^3=-(y-3)^3\Rightarrow x-4=-(y-3)\Rightarrow x=7-y$
Thế vào pt (1) hoặc (2) rồi tìm x,y e nhé

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