Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)Với x \(\ne\)-1
Ta có: x2 + x = 0
=> x(x + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-1\left(ktm\right)\end{cases}}\)
Với x = 0 => A = \(\frac{0-3}{0+1}=-3\)
b) Ta có: B = \(\frac{3}{x-3}+\frac{6x}{9-x^3}+\frac{x}{x+3}\)
B = \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
B = \(\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x+3}{x-3}\)
c) Với x \(\ne\)\(\pm\)3; x \(\ne\)-1
Ta có: P = AB = \(\frac{x-3}{x+1}\cdot\frac{x+3}{x-3}=\frac{x+3}{x+1}=\frac{\left(x+1\right)+2}{x+1}=1+\frac{2}{x+1}\)
Để P \(\in\)Z <=> 2 \(⋮\)x + 1
<=> x + 1 \(\in\)Ư(2) = {1; -1; 2; -2}
<=> x \(\in\){0; -2; 1; -3}
a) ĐK: \(x\ne0,x\ne\pm3\)
\(A=\left(\frac{x-3}{x^2-9}+\frac{1}{x+3}\right)\div\frac{x}{x+3}\)
\(=\left(\frac{1}{x+3}+\frac{1}{x+3}\right)\div\frac{x}{x+3}\)
\(=\frac{2}{x+3}\times\frac{x+3}{x}=\frac{2}{x}\)
b) \(\left|A\right|=\left|\frac{2}{x}\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}=3\\\frac{2}{x}=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)(thỏa mãn)
\(a.2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(t/mđk\right)\\x=3\left(loại,kot/mđk\right)\end{cases}}\)
\(Thay:x=0\left(t/mđk\right)\Leftrightarrow A=\frac{x-3}{x+3}\Rightarrow\frac{0-3}{0+3}=-\frac{3}{3}=-1\left(t/mđk\right)\)
\(\text{a, ĐKXĐ: }\hept{\begin{cases}x+3\ne0\\x-3\ne0\\3x^2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\mp3\\x\ne0\end{cases}}\)
\(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left[\frac{\left(3-x\right)\left(x+3\right)^2}{\left(x+3\right)\left(x+3\right)\left(x-3\right)}+\frac{x}{x+3}\right]\cdot\frac{x+3}{3x^2}\)
\(=\frac{x-x-3}{x+3}\cdot\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
b, với x=\(-\frac{1}{2}\)ta có:
\(A=-\frac{1}{\left(-\frac{1}{2}\right)^2}=-4\)
c, Để A<0 thì \(-\frac{1}{x^2}< 0\text{ mà }x^2>0\left(\text{vì x khác 0 ĐKXĐ}\right)\)
Với x khác 0 thì thỏa mãn!
a) ĐKXĐ: \(x\ne\pm3\)
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\frac{\left(3-x\right)\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=\frac{3\left(3-x\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x+3}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left[\frac{\left(3-x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right]:\frac{3x^2}{x+3}\)
\(A=\left(\frac{9-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}\times\frac{x+3}{3x^2}\)
\(A=\frac{-1}{x^2}\)
Ta có :\(x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(L\right)\\x=2\left(tm\right)\end{cases}}\)
\(\Rightarrow A=\frac{-1}{2^2}\)
\(A=\frac{-1}{4}\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\x\ne0\end{cases}}\)
a) \(B=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\left(\frac{3-x}{x+3}\cdot\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\frac{\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{3x^2}\)
\(\Leftrightarrow B=-\frac{x+3}{3x^2}\)
b) Khi \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=3\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x=1\)
\(\Leftrightarrow B=-\frac{1+3}{3.1^2}=-\frac{4}{3.}\)
c) Để B > 0
\(\Leftrightarrow-\frac{x+3}{3x^2}>0\)
\(\Leftrightarrow\frac{x+3}{3x^2}< 0\)
\(\Leftrightarrow x+3< 0\) (Do 3x2 > 0; loại giá trị = 0)
\(\Leftrightarrow x< -3\)
Vậy để \(B>0\Leftrightarrow x< -3\)
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left[\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)
\(=\left[\frac{-\left(x-3\right)\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)^2}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)
\(=\left(-1+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)
\(=\frac{-x-3+x}{x+3}.\frac{x+3}{3x^2}=\frac{-3}{x+3}.\frac{x+3}{3x^2}=\frac{-1}{x^2}\)
b ) Để \(A=-\frac{1}{x^2}< 0\forall x\ne0\)
Vậy \(x\ne0\) thì \(A< 0\)
\(A=\left(\frac{-\left(x-3\right)}{\left(x+3\right)}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right).\left(\frac{x+3}{3x^2}\right)\)
\(=\left(-1+\frac{x}{x+3}\right)\left(\frac{x+3}{3x^2}\right)=\frac{-3}{\left(x+3\right)}.\frac{\left(x+3\right)}{3x^2}=\frac{-1}{x^2}\)
\(A< 0\Rightarrow\frac{-1}{x^2}< 0\Rightarrow-1< 0\) (luôn đúng)
Vậy \(x\ne0;x\ne\pm3\) thì \(A< 0\)