\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)
1. Rút gọn A
2. Tìm x để A nhỏ hơn -1
3. Với GT nào của x thì A nhận GT nguyên
K
Khách
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S
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
CC
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 \(\Leftrightarrow x=\pm1\)
MN
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 \(A\inℤ\Leftrightarrow x\inƯ\left(3\right)\)(\(x\neℤ\))
MN
10 tháng 2 2020
Bạn sửa cho mik dòng cuối :
\(x\ne Z\)thành \(x\notin Z\)nhé !
21 tháng 4 2020
a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
= \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)
<=> \(\frac{x^2+1}{x+1}+1>0\)
<=> \(\frac{x^2+x+2}{x+1}>0\)
Vì x2 + x + 2 >0 \(\forall x\)
=> A > 0 <=> x + 1 > 0 <=> x > -1