a: ĐểA nguyên thì x^2+2x+x+2-3 chia hết cho x+2

=>-3 chia hết cho x+2

=>x+2 thuộc {1;-1;3;-3}

=>x thuộc {-1;-3;1;-5}

b: B nguyên khi x^2+x+3 chia hết cho x+1

=>3 chia hết cho x+1

=>x+1 thuộc {1;-1;3;-3}

=>x thuộc {0;-2;2;-4}

Bạn xem thử tại đây:

https://hoc24.vn/cau-hoi/cho-bieu-thucm-dfrac2a2-a-dfrac4a2a2-4-dfrac2-a2aa-rut-gon-mb-tinh-gia-tri-cua-m-khi-a13c-tim-a-z-de-m-la-so-nguyen-chia-het-cho-4.7975358921144

a,bn viết đúng đề xíu nhé \(\dfrac{\sqrt{a}+2}{\sqrt{a+3}}\) sửa \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}\)

đk: \(a\ge0,a\ne4\)

=>\(P=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{1}{\sqrt{a}-2}\)

\(=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)\(=\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)

b, \(P=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=1+\dfrac{-2}{\sqrt{a}-2}\) nguyên\(< =>\sqrt{a}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(=>a\in\left\{9;1;16;0\right\}\)(TM)

a) P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\left(ĐKXĐ:a\ge0;a\ne4\right)\)

P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\dfrac{1}{\sqrt{a}-2}\)

P = \(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

P = \(\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

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

P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) 

b) Ta có: P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) = 1 - \(\dfrac{2}{\sqrt{a}-2}\)

Để \(P\in Z\) <=> 1 - \(\dfrac{2}{\sqrt{a}-2}\) \(\in Z\) <=> \(\sqrt{a}-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Ta có bảng sau: 

Vậy để \(P\in Z\) thì  \(a\in\left\{0;1;9;16\right\}\)

a,A = \(\dfrac{3}{x-1}\)

A \(\in\) Z \(\Leftrightarrow\)  3 ⋮ \(x-1\)  ⇒ \(x-1\) \(\in\) { -3; -1; 1; 3}

                                    \(x\) \(\in\) { -2; 0; 2; 4}

b, B =  \(\dfrac{x-2}{x+3}\)  

B \(\in\) Z \(\Leftrightarrow\) \(x-2\) \(⋮\) \(x+3\) ⇒ \(x+3-5\) \(⋮\) \(x+3\)

                                   ⇒               5  \(⋮\) \(x+3\)

                                  \(x+3\) \(\in\){ -5; -1; 1; 5}

                                  \(x\) \(\in\) { -8; -4; -2; 2}

a.\(A=\dfrac{3}{x-1}\)có giá trị là 1 số nguyên khi \(3\) ⋮ \(x-1.\)

\(\Rightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}.\)

Ta có bảng:

Vậy \(x\in\left\{-2;0;2;4\right\}.\)

b.\(B=\dfrac{x-2}{x+3}\)có giá trị là 1 số nguyên khi \(x-2\) ⋮ \(x+3.\)

\(\Rightarrow\left(x+3\right)-5⋮x+3.\) 

Mà x+3 ⋮ x+3 \(\Rightarrow\) Ta cần: \(-5⋮x+3\Rightarrow x+3\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}.\) 
Ta có bảng:

Vậy \(x\in\left\{-8;-4;-2;2\right\}.\)
 

\(\dfrac{2}{3}A=\dfrac{2}{3}-\left(\dfrac{2}{3}\right)^2+\left(\dfrac{2}{3}\right)^3-...+\left(\dfrac{2}{3}\right)^{2019}-\left(\dfrac{2}{3}\right)^{2020}\)

=>\(\dfrac{5}{3}A=1-\left(\dfrac{2}{3}\right)^{2020}=1-\dfrac{2^{2020}}{3^{2020}}=\dfrac{3^{2020}-2^{2020}}{3^{2020}}\)

=>\(A=\dfrac{3^{2020}-2^{2020}}{3^{2020}}:\dfrac{5}{3}=\dfrac{3^{2020}-2^{2020}}{5\cdot3^{2019}}\) ko là số nguyên

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

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 \(a\in\left\{1;16;25;49\right\}\)