A = \(6\)

bạn có thể giải chi tiết giúp mình đc ko

A nguyên

=>3căn x+9-7 chia hết cho 2 căn x+3

=>2căn x+3=7

=>x=4

a) ĐKXĐ: x∉{2;-2}

b) Ta có: \(A=\dfrac{x}{x-2}+\dfrac{2-x}{x+2}+\dfrac{12-10x}{x^2-4}\)

\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x-x^2+4x-4+12-10x}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{-4x+8}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{-4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{-4}{x+2}\)

c) Để \(A=\dfrac{2}{3}\) thì \(\dfrac{-4}{x+2}=\dfrac{2}{3}\)

\(\Leftrightarrow x+2=\dfrac{-4\cdot3}{2}=-\dfrac{12}{2}=-6\)

hay x=-6-2=-8(nhận)

Vậy: Để \(A=\dfrac{2}{3}\) thì x=-8

d) Để A nguyên thì \(-4⋮x+2\)

\(\Leftrightarrow x+2\inƯ\left(-4\right)\)

\(\Leftrightarrow x+2\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow x\in\left\{-1;-3;0;-4;2;-6\right\}\)(nhận)

Vậy: Để A nguyên thì \(x\in\left\{-1;-3;0;-4;2;-6\right\}\)

Lời giải:
\(\frac{2-3\sqrt{x}}{\sqrt{x}+3}=\frac{11-3(\sqrt{x}+3)}{\sqrt{x}+3}=\frac{11}{\sqrt{x}+3}-3\)

Để biểu thức đã cho nguyên thì $\frac{11}{\sqrt{x}+3}$ nguyên

Đặt $\frac{11}{\sqrt{x}+3}=t$ thì hiển nhiên $t>0$ do cả tử và mẫu đều dương.

Mà: $\sqrt{x}\geq 0\Rightarrow t=\frac{11}{\sqrt{x}+3}\leq \frac{11}{3}<4$

$\Rightarrow 0< t< 4$. Mà $t$ nguyên nên $t\in \left\{1; 2; 3\right\}$

$\sqrt{x}=\frac{11}{t}-3$. Để $x$ nguyên thì $t$ là ước của $11$

$\Rightarrow t=1$

$\sqrt{x}=\frac{11}{1}-3=8\Leftrightarrow x=64$

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

\(=\dfrac{-2\left(x-3\right)}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{x-1}{x-3}=\dfrac{-2}{x+1}\)

b: Để P<0 thì x+1>0

hay x>-1

c: Để Q=(-2x)/(x+1) là số nguyên thì \(-2x-2+2⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{0;-2;-3\right\}\)

a: \(P=\dfrac{x}{x+3}-\dfrac{x^2-5x-6}{\left(x-3\right)\left(x+3\right)}+\dfrac{3}{x-3}\)

\(=\dfrac{x^2-3x-x^2+5x+6+3x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{5}{x-3}\)

a, ĐKXĐ:\(\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x^2+x-6\ne0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

b, \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

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

\(=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{x-4}{x-2}\)

 \(c,A=\dfrac{-3}{4}\\ \Leftrightarrow\dfrac{x-4}{x-2}=\dfrac{-3}{4}\\ \Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\\ \Leftrightarrow4x-16x=-3x+6\\ \Leftrightarrow4x-16x+3x-6=0\\ \Leftrightarrow7x-22=0\\ \Leftrightarrow x=\dfrac{22}{7}\)

d, \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)

Để \(A\in Z\Rightarrow\dfrac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Ta có bảng:
 

Vậy \(x\in\left\{0;1;3;4\right\}\)

a)x khác -3 và x khác 2 =)

a: ĐXKĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

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

c: Để A=-3/4 thì x-4/x-2=-3/4

=>4x-16=-3x+6

=>7x=22

hay x=22/7