Question

[Min Suga]

Cho A = \(\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}+1}{3-\sqrt{a}}\)

a) Rút gọn A 

b) Tìm a để A <1

c) Tìm aϵ Z để A ϵ Z

Answer 1

[Akai Haruma]

Lời giải:ĐK: $a\geq 0; a\neq 9; a\neq 4$

a) 

\(A=\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{\sqrt{a}+3}{\sqrt{a}-2}+\frac{2\sqrt{a}+1}{\sqrt{a}-3}\)

\(\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{(\sqrt{a}+3)(\sqrt{a}-3)}{(\sqrt{a}-2)(\sqrt{a}-3)}+\frac{(2\sqrt{a}+1)(\ \sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}\)

\(=\frac{2\sqrt{a}-9-(a-9)+(2a-3\sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{a-\sqrt{a}-2}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{(\sqrt{a}-2)(\sqrt{a}+1)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)

b) Để \(A< 1\Leftrightarrow \frac{\sqrt{a}+1}{\sqrt{a}-3}<1\Leftrightarrow 1+\frac{4}{\sqrt{a}-3}<1\)

\(\Leftrightarrow \frac{4}{\sqrt{a}-3}< 0\Leftrightarrow \sqrt{a}-3< 0\Leftrightarrow 0\leq a< 9\)

Kết hợp ĐKXĐ: suy ra $0\leq a< 9; a\neq 4$

c) Với $a$ nguyên,  \(A=1+\frac{4}{\sqrt{a}-3}\in\mathbb{Z}\Leftrightarrow 4\vdots \sqrt{a}-3\)

$\Rightarrow \sqrt{a}-3\in\left\{\pm 1; \pm 2;\pm 4\right\}$

$\Rightarrow a\in\left\{4;16; 1;25; 49\right\}$

Kết hợp ĐKXĐ suy ra $a\in\left\{16;1;25;49\right\}$

Answer 2

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{4;9\right\}\end{matrix}\right.\)

a) Ta có: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}+1}{3-\sqrt{a}}\)

\(=\dfrac{\left(2\sqrt{a}-9\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}-\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}+\dfrac{\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)

\(=\dfrac{2\sqrt{a}-9-\left(a-9\right)+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{2a-\sqrt{a}-11-a+9}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{a-2\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)+\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)

b) Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-1< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-\dfrac{\sqrt{a}-3}{\sqrt{a}-3}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\)

\(\Leftrightarrow\dfrac{4}{\sqrt{a}-3}< 0\)

mà 4>0

nên \(\sqrt{a}-3< 0\)

\(\Leftrightarrow\sqrt{a}< 3\)

hay a<9

Kết hợp ĐKXĐ, ta được:

\(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)

Vậy: Để A<1 thì \(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)

c) Để A nguyên thì \(\sqrt{a}+1⋮\sqrt{a}-3\)

\(\Leftrightarrow\sqrt{a}-3+4⋮\sqrt{a}-3\)

mà \(\sqrt{a}-3⋮\sqrt{a}-3\)

nên \(4⋮\sqrt{a}-3\)

\(\Leftrightarrow\sqrt{a}-3\inƯ\left(4\right)\)

\(\Leftrightarrow\sqrt{a}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

mà \(\sqrt{a}-3\ge-3\forall a\) thỏa mãn ĐKXĐ

nên \(\sqrt{a}-3\in\left\{1;-1;2;-2;4\right\}\)

\(\Leftrightarrow\sqrt{a}\in\left\{4;2;5;1;7\right\}\)

\(\Leftrightarrow a\in\left\{16;4;25;1;49\right\}\)

Kết hợp ĐKXĐ, ta được: \(a\in\left\{1;16;25;49\right\}\)

Vậy: Để A nguyên thì \(a\in\left\{1;16;25;49\right\}\)