3/4×4/6×5/8…2010/4018=1005×16^1-x tìm x tìm cặp số nguyên x y thỏa mãn x^2+xy-2y-3x=3

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1. Đáp án: a.$x=503$

                b.$(x,y)\in\{(3,3), (7,-5), (1,-5), (-3,3)\}$

   Giải thích các bước giải:

   a.Ta có:

   $\dfrac34\cdot \dfrac46\cdot 58…\cdot\dfrac{2010}{4018}=1005\cdot 16^{1-x}$

   $\to \dfrac{3\cdot 4\cdot 5\cdots2010}{4\cdot 6\cdot 8\cdots4018}=1005\cdot 16^{1-x}$

   $\to \dfrac{3\cdot 4\cdot 5\cdots2010}{(2\cdot 2)\cdot (2\cdot 3)\cdot (2\cdot 4)\cdots(2\cdot 2009)}=1005\cdot 16^{1-x}$

   $\to \dfrac{3\cdot 4\cdot 5\cdots2010}{2^{2008}\cdot (2\cdot 3\cdot 4\cdots2009)}=1005\cdot 16^{1-x}$

   $\to \dfrac{2010}{2^{2008}\cdot 2}=1005\cdot 16^{1-x}$

   $\to \dfrac{1005}{2^{2008}}=1005\cdot 16^{1-x}$

   $\to \dfrac{1}{2^{2008}}=16^{1-x}$

   $\to 2^{-2008}=(2^4)^{1-x}$

   $\to 2^{-2008}=2^{4(1-x)}$

   $\to -2008=4(1-x)$

   $\to 1-x=-502$

   $\to x=503$

   b.Ta có:

   $x^2+xy-2y-3x=3$

   $\to xy-2y=-(x^2-3x-3)$

   $\to y(x-2)=-(x^2-2x-x+2-5)$

   $\to y(x-2)=-(x(x-2)-(x-2)-5)$

   $\to y(x-2)=-((x-1)(x-2)-5)$

   $\to y(x-2)=-(x-1)(x-2)+5$

   $\to 5\quad\vdots\quad x-2$

   $\to x-2\in\{1,5,-1,-5\}$

   $\to x\in\{3, 7, 1, -3\}$

   Mà $y(x-2)=-(x-1)(x-2)+5\to y=-(x-1)+\dfrac{5}{x-2}$

   $\to y\in\{3,-5,-5, 3\}$

   $\to (x,y)\in\{(3,3), (7,-5), (1,-5), (-3,3)\}$

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