1.1
\(A=\dfrac{\sqrt x+7}{\sqrt x+2}(x\ge 0)\\=\dfrac{\sqrt x+2+5}{\sqrt x+2}\\=1+\dfrac{5}{\sqrt x+2}\)
Để \(A\in\Bbb Z\)
\(→1+\dfrac{5}{\sqrt x+2}\in\Bbb Z\\→\dfrac{5}{\sqrt x+2}\in \Bbb Z\\→5:\sqrt x+2\\→\sqrt x+2\in Ư(5)=\{±1;±5\}\)
mà \(\sqrt x+2≥2\)
\(→\sqrt x+2=5\\↔\sqrt x=3\\↔x=9(TM)\)
Vậy \(x=9\) thì \(A\in\Bbb Z\)
2.1
\(A=\dfrac{\sqrt x+7}{\sqrt x+2}(x\ge 0)\\=\dfrac{\sqrt x+2+5}{\sqrt x+2}\\=1+\dfrac{5}{\sqrt x+2}\)
Vì \(A∈\Bbb Z→1+\dfrac{5}{\sqrt x+2}\in\Bbb Z\)
\(→\dfrac{5}{\sqrt x+2}\in\Bbb Z\\→5\vdots \sqrt x+2\\→\sqrt x+2\in Ư(5)=\{±1;±5\}\)
mà \(\sqrt x+2\ge 2\)
\(→\sqrt x+2=5\\↔\sqrt x=3\\↔x=9(TM)\)
Vậy \(x=9\)
3.1
\(A=\dfrac{x+\sqrt x+3}{\sqrt x+1}\\=\dfrac{\sqrt x(\sqrt x+1)+3}{\sqrt x+1}\\=\sqrt x+\dfrac{3}{\sqrt x+1}\\=(\sqrt x+1)+\dfrac{3}{\sqrt x+1}-1\)
Áp dụng BĐT Cô-si với hai số dương \(\sqrt x+1,\dfrac{3}{\sqrt x+1}\)
\(→A≥2\sqrt{(\sqrt x+1).\dfrac{3}{\sqrt x+1}}-1\\→A≥2\sqrt 3-1\)
\(→\min A=2\sqrt 3-1\)
\(→\) Dấu "=" xảy ra khi \(\sqrt x+1=\dfrac{3}{\sqrt x+1}\)
\(↔(\sqrt x+1)^2=3\\↔\sqrt x+1=\sqrt 3(vì\,\,\sqrt x+1>0)\\↔\sqrt x=\sqrt 3-1\\↔x=4-2\sqrt 3\)
Vậy \(\min A=2\sqrt 3-1\) khi \(x=4-2\sqrt 3\)
