Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right):\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{6}=\dfrac{-1}{x-2}\)
b: |x|=1/2 khi x=1/2 hoặc x=-1/2
Khi x=1/2 thì \(A=\dfrac{-1}{\dfrac{1}{2}-2}=-1:\dfrac{-3}{2}=\dfrac{2}{3}\)
Khi x=-1/2 thì \(A=\dfrac{-1}{-\dfrac{1}{2}-2}=-1:\dfrac{-5}{2}=\dfrac{2}{5}\)
c: Để A=2 thì x-2=-1/2
hay x=3/2
d:Để A<0 thì x-2>0
hay x>2
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
Lời giải:
ĐKXĐ: \(x\neq \left\{2;\pm 3\right\}\)
a) Ta có:
\(P=\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x(x-3)}{(x-3)(x+3)}-1\right):\left(\frac{(3-x)(3+x)}{(x-2)(x+3)}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x}{x+3}-1\right):\left(\frac{3-x}{x-2}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\frac{x-(x+3)}{x+3}:\left(-\frac{x-2}{x+3}\right)=\frac{-3}{x+3}.\frac{x+3}{-(x-2)}=\frac{3}{x-2}\)
b) \(x^3-3x+2=0\)
\(\Leftrightarrow (x^3-x)-2(x-1)=0\)
\(\Leftrightarrow x(x-1)(x+1)-2(x-1)=0\)
\(\Leftrightarrow (x-1)(x^2+x-2)=0\)
\(\Leftrightarrow (x-1)[(x^2-1)+(x-1)]=0\)
\(\Leftrightarrow (x-1)^2(x+2)=0\) \(\Leftrightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Với \(x=1\Rightarrow P=\frac{3}{1-2}=-3\)
Với \(x=-2\Rightarrow P=\frac{3}{-2-2}=\frac{-3}{4}\)
c)
\(P=\frac{3}{x-2}\in\mathbb{Z}\Leftrightarrow 3\vdots x-2\)
\(\Leftrightarrow x-2\in \text{Ư}(3)\Rightarrow x-2\in\left\{\pm 1; \pm 3\right\}\)
\(\Leftrightarrow x\in \left\{3,1,5,-1\right\}\)
Do \(x\neq 3\Rightarrow x\in \left\{-1,1,5\right\}\)
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
a.
ĐKXĐ: \(x\ne2\)
b.
\(P=\left(\dfrac{2x}{x-2}+\dfrac{x}{2-x}\right):\dfrac{x^2+1}{x-2}\)
\(=\left(\dfrac{2x}{x-2}-\dfrac{x}{x-2}\right)\cdot\dfrac{x-2}{x^2+1}\)
\(=\dfrac{x}{x-2}\cdot\dfrac{x-2}{x^2+1}=\dfrac{x}{x^2+1}\)
c.
\(x=-1\Rightarrow P=-\dfrac{1}{\left(-1\right)^2+1}=-\dfrac{1}{2}\)
d.
\(P=\dfrac{x}{x^2+1}\cdot\dfrac{x^2+1}{x}-\dfrac{1}{P}\ge1-\dfrac{1}{P}\)
\(\Rightarrow\dfrac{P^2+1}{P}\ge1\)
\(\Rightarrow P^2+1\ge P\) \(\Rightarrow P\left(P-1\right)\ge1\)
\(\Rightarrow P\ge2\)
Dấu "=" khi x = ...................
Bài 2:
a: \(M=\dfrac{3x+1-2x-2}{\left(3x-1\right)\left(3x+1\right)}:\dfrac{3x+1-3x}{x\left(3x+1\right)}\)
\(=\dfrac{x-1}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{x\left(3x+1\right)}{1}=\dfrac{x\left(x-1\right)}{3x-1}\)
b: Để M=0 thì x(x-1)=0
=>x=1(nhận) hoặc x=0(loại)
c: \(P=M\cdot\left(3x-1\right)=x\left(x-1\right)=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/2
Bài 1:
a: ĐKXĐ: x<>1/3; x<>-1/3
b: \(M=\left(\dfrac{-3x}{3x-1}+\dfrac{2x}{3x+1}\right)\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)
\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)
\(=\dfrac{-3x^2-5x}{\left(3x+1\right)}\cdot\dfrac{1}{2\left(3x^2+5\right)}=\dfrac{-3x^2-5x}{2\left(3x+1\right)\left(3x^2+5\right)}\)
Bài 2:
\(P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)
\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)
\(=\dfrac{5}{x-5}+\dfrac{x}{5-x}=-1\)
\(\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{4x-2}{x+1}-\dfrac{1-x^2+3x}{3x}=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}.\dfrac{x+1}{4x-2}-\dfrac{1-x^2+3x}{3x}=\dfrac{-8x^2+2}{3x\left(4x-2\right)}-\dfrac{1-x^2+3x}{3x}=\dfrac{-2x-1-1+x^2-3x}{3x}=\dfrac{x^2-5x-2}{3x}\)
a/ rút gọn ra là M= \(\dfrac{x^2-5x-2}{3x}\)
b/ thay x=123 thì M=\(\dfrac{14512}{369}\)