Đáp án: Bài 2: $C\ge 34$
Giải thích các bước giải:
Bài 1:
Ta có:
$B=111\cdots11-22\cdots22$
$\to B=\dfrac19\cdot 99\cdots99-\dfrac29\cdot 99\cdots99$
$\to B=\dfrac19\left(10^{4036}-1\right)-\dfrac29\left(10^{2018}-1\right)$
$\to B=\dfrac19\left(10^{4036}-1-2\left(10^{2018}-1\right)\right)$
$\to B=\dfrac19\left(\left(10^{2018}\right)^2-2\cdot10^{2018}+1\right)$
$\to B=\dfrac19\left(10^{2018}-1\right)^2$
$\to B=\left(\dfrac{10^{2018}-1}{3}\right)^2$
Ta có : $10^{2018}-1\quad\vdots\quad 10-1=9$
$\to 10^{2018}-1\quad\vdots\quad 3$
$\to B=\left(\dfrac{10^{2018}-1}{3}\right)^2$ là số chính phương
Bài 2:
Ta có:
$C=5x+6y+\dfrac{10}x+\dfrac{14}y$
$\to C=\left(\dfrac52x+\dfrac52y\right)+\left(\dfrac52x+\dfrac{10}x\right)+\left(\dfrac72y+\dfrac{14}y\right)$
$\to C=\dfrac52\left(x+y\right)+\left(\dfrac52x+\dfrac{10}x\right)+\left(\dfrac72y+\dfrac{14}y\right)$
$\to C\ge \dfrac52\cdot 4+2\sqrt{\dfrac52x\cdot\dfrac{10}x}+2\sqrt{\dfrac72y\cdot\dfrac{14}y}$
$\to C\ge 34$
Dấu = xảy ra khi $x=y=2$
