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14 tháng 12 2018
a) P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}}\)
Vậy P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{\left(x-5\right)\left(x+5\right)2}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
Có: \(P=0\)
\(\Rightarrow P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=0\Leftrightarrow x\left(x^2+4x-5\right)=0\Leftrightarrow x^2+4x-5=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy \(P=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
3 tháng 1 2019
a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+5x^2-x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)
\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x-1\right)}{2x}\)
\(P=\frac{x-1}{2}\)
c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )
Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )
d) Để P > 0 thì \(\frac{x-1}{2}>0\)
Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)
Để P < 0 thì \(\frac{x-1}{2}< 0\)
Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)
KT
14 tháng 4 2018
a) ĐKXĐ: \(x\ne\pm2\)
\(A=\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x-2x+4}{x^2-4}\)\(=\frac{x^2+4}{x^2-4}\)
b) \(A>0\) \(\Rightarrow\)\(\frac{x^2+4}{x^2-4}>0\)
Mà \(x^2+4>0\) \(\Rightarrow\)\(x^2-4>0\)
\(\Rightarrow\)\(x^2>4\)
Nếu x dương thì \(x>\sqrt{4}=2\)
Nếu x âm thì \(x< \sqrt{4}=2\)
20 tháng 8 2022
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(B=\left(\dfrac{x}{x^2-4}-\dfrac{2}{x^2-2x}+\dfrac{1}{x+2}\right):\left(\dfrac{10-x^2}{x+2}+x-2\right)\)
\(=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{x\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-2x}{x\left(x-2\right)\left(x+2\right)}\right):\dfrac{10-x^2+x^2-4}{x+2}\)
\(=\dfrac{x^2-2x-4+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{2x^2-4x-4}{x\left(x-2\right)}\cdot\dfrac{1}{6}\)
\(=\dfrac{x^2-2x-2}{x\left(x-2\right)}\)
b: Để B=0 thì \(x^2-2x-2=0\)
hay \(x\in\left\{1+\sqrt{3};1-\sqrt{3}\right\}\)
HL
13 tháng 7 2017
a) \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
( ĐKXĐ : \(x\ne0,x\ne-5\) )
\(B=\dfrac{\left(x^2+2x\right).x}{2x\left(x+5\right)}+\dfrac{\left(x-5\right).2\left(x+5\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3+2x^2+2x^2+10x-10x-50+50-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^3-x^2+5x^2-5x}{2x\left(x+5\right)}\)
\(B=\dfrac{x^2\left(x-1\right)+5x\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{\left(x-1\right)\left(x+5\right)x}{2x\left(x+5\right)}\)
\(B=\dfrac{x-1}{2}\)
Câu b và c dễ vì đã có kết quả rút gọn rồi :)
2 tháng 5 2018
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
7 tháng 12 2022
a: ĐKXĐ: x<>0; x<>-5
b: \(P=\dfrac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+50-5x+2x^2-50}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
Để P=0 thì x-1=0
=>x=1
c: Để P=-1/4 thì x-1/2=-1/4
=>x-1=-1/2
=>x=1/2
19 tháng 8 2017
a) điều kiện \(x\ne\pm2\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{x^2-4}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x-6\right)}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{1}{x-2}.\dfrac{3x-2x^2-6}{1}=\dfrac{3x-2x^2-6}{x-2}\)
b) ta có : \(3x-2x^2-6=-2x^2+3x-6=-\left(2x^2-3x+6\right)\)
\(=\left(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3}{2\sqrt{2}}+\left(\dfrac{3}{2\sqrt{2}}\right)^2\right)+\dfrac{39}{8}\)
\(=\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\ge\dfrac{39}{8}>0\)
\(\Rightarrow A\le0\) \(\Leftrightarrow x-2\le0\) (mà đk : \(x\ne2\) \(\Rightarrow x-2\ne0\))
vậy \(A\le0\Leftrightarrow A< 0\) \(\Leftrightarrow x-2< 0\Leftrightarrow x< 2\) vậy \(x< 2\)
Để \(B< 0\) mà 30>0\(\Rightarrow x-2< 0\Rightarrow x< 2\)
x\(\le\)31 (\(\forall\)x)
và x\(\notin\left\{-4,4\right\}\) thì B<0