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 $y(x-2)=-(x-1)(x-2)+5\to y=-(x-1)+\dfrac{5}{x-2}$
$\to y\in\{3,-5,-5, 3\}$
$\to (x,y)\in\{(3,3), (7,-5), (1,-5), (-3,3)\}$
