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23 tháng 12 2019

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-3\\x\ne3\end{cases}}\)

\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)\(=\left[\frac{1}{3}+\frac{3}{x\left(x-3\right)}\right]:\left(\frac{-x^2}{3x^2-27}+\frac{1}{x+3}\right)\)

\(=\left[\frac{x\left(x-3\right)}{3x\left(x-3\right)}+\frac{9}{3x\left(x-3\right)}\right]:\left[\frac{-x^2}{3\left(x^2-9\right)}+\frac{1}{x+3}\right]\)

\(=\frac{x^2-3x+9}{3x\left(x-3\right)}:[\frac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\frac{3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}]\)

\(=\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{-x^2+3x-9}{3\left(x-3\right)\left(x+3\right)}\)\(=\frac{x^2-3x+9}{3x\left(x-3\right)}.\frac{3\left(x-3\right)\left(x+3\right)}{-\left(x^2-3x+9\right)}=\frac{x+3}{-x}=\frac{-x-3}{x}=-1-\frac{3}{x}\)

b) \(A< -1\)\(\Leftrightarrow-1-\frac{3}{x}< -1\)\(\Leftrightarrow\frac{-3}{x}< 0\)

mà \(-3< 0\)\(\Rightarrow x>0\)và \(x\ne3\)

Vậy \(A< -1\Leftrightarrow\hept{\begin{cases}x>0\\x\ne3\end{cases}}\)

c) Vì \(-1\inℤ\)\(\Rightarrow\)Để A nguyên thì \(\frac{3}{x}\inℤ\)\(\Rightarrow3⋮x\)

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

So sánh với ĐKXĐ \(\Rightarrow x=\pm3\)loại

Vậy A nguyên \(\Leftrightarrow x=\pm1\)

20 tháng 12 2018

\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

\(=\left(\frac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

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

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

b) \(Voix>0\Rightarrow P\ne\varnothing\)(mk ko chac)

c) \(P\inℤ\Leftrightarrow x+3⋮x-3\Leftrightarrow x-3\in\left\{-1;-2;-3;-6;1;2;3;6\right\}\) 

sau do tinh

cau nay la toan lp 8 nha

20 tháng 12 2018

P= O/ nha

10 tháng 2 2020

Đề sai ạ ! Sửa lại nhé : 

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)

\(\Leftrightarrow A=\frac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\frac{-x^2}{3\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right)\)

\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{-x^2+3\left(x-3\right)}{3\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{x^2-3x+9}{3x\left(x-3\right)}.\frac{3\left(x-3\right)\left(x+3\right)}{-x^2+3x-9}\)

\(\Leftrightarrow A=\frac{-\left(x+3\right)}{x}\)

b) Để \(A\inℤ\)

\(\Leftrightarrow-\left(x+3\right)⋮x\)

\(\Leftrightarrow-x-3⋮x\)

\(\Leftrightarrow3⋮x\)

\(\Leftrightarrow x\inƯ\left(3\right)\)

Vậy để \(A\inℤ\Leftrightarrow x\inƯ\left(3\right)\)(\(x\neℤ\))

10 tháng 2 2020

Bạn sửa cho mik dòng cuối :

\(x\ne Z\)thành \(x\notin Z\)nhé !

7 tháng 9 2019

PLEASE HELP ME !!!

11 tháng 10 2020

a) Ta có: \(3x+2\sqrt{3x}+4=\left(\sqrt{3x}+1\right)^2+3>0;1+\sqrt{3x}>0,\forall x\ge0\), nên đk để A có nghĩa là

\(\left(\sqrt{3x}\right)^3-8-\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)\ne0;x\ge0\Leftrightarrow\sqrt{3x}\ne2\Leftrightarrow0\le x\ne\frac{4}{3}\)

A=\(\left(\frac{6x+4}{\left(\sqrt{3x}\right)^3-2^3}-\frac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\frac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)

\(=\left(\frac{6x+4-\left(\sqrt{3x}-2\right)\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\left(3x-\sqrt{3x}+1-\sqrt{3x}\right)\)

\(=\left(\frac{3x+4+2\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\left(3x-2\sqrt{3x}+1\right)\)

\(=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\left(0\le x\ne\frac{4}{3}\right)\)

b) \(A=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}=\frac{\left(\sqrt{3x}-2\right)^2+2\left(\sqrt{3x}-2\right)+1}{\sqrt{3x}-2}=\sqrt{3x}+\frac{1}{\sqrt{3x}-2}\)

Với \(x\ge0\), để A là số nguyên thì \(\sqrt{3x}-2=\pm1\Leftrightarrow\orbr{\begin{cases}\sqrt{3x}=3\\\sqrt{3x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=9\\3x=1\end{cases}\Leftrightarrow}x=3}\)  (vì \(x\in Z;x\ge0\))

Khi đó A=4

25 tháng 11 2019

a) A = \(\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\left(\frac{1}{1-x}-1\right)\)

A = \(\frac{3x^2+3x-3}{x^2+2x-x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\left(\frac{1-1+x}{1-x}\right)\)

A = \(\frac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\frac{x}{1-x}\)

A = \(\frac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}-\frac{x+1}{x+2}-\frac{x-2}{x-1}\)

A = \(\frac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

A = \(\frac{3x^2+3x-3-x^2+1-x^2+4}{\left(x-1\right)\left(x+2\right)}\)

A = \(\frac{x^2+3x+2}{\left(x-1\right)\left(x+2\right)}\)

A = \(\frac{x^2+2x+x+2}{\left(x-1\right)\left(x+2\right)}\)

A = \(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

A = \(\frac{x+1}{x-1}\) (Đk: \(x-1\ge0\) => x \(\ge\)1)

b) Ta có: A = \(\frac{x+1}{x-1}=\frac{\left(x-1\right)+2}{x-1}=1+\frac{2}{x-1}\)

Để A \(\in\)Z <=> 2 \(⋮\)x - 1

<=> x - 1 \(\in\)Ư(2) = {1; -1; 2; -2}

<=> x \(\in\){2; 0; 3; -1}

c) Ta có: A < 0

=> \(\frac{x+1}{x-1}< 0\)

=> \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)

=> \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(loại) hoặc \(\hept{\begin{cases}x>-1\\x< 1\end{cases}}\) 

=> -1 < x < 1

25 tháng 11 2019

Edogawa Conan

Thiếu dòng đầu  \(ĐKXĐ:\hept{\begin{cases}x\ne1\\x\ne-2\\x\ne0\end{cases}}\)