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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 \(x-1< 0\Leftrightarrow x< 1\)
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: \(x\in\left\{-5;3;-3\right\}\)
\(A=\dfrac{-3\left(x+5\right)}{\left(x+5\right)^2}:\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-3}{x+5}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-3\left(x+3\right)}\)
\(=\dfrac{x-3}{x+5}\)
b: Để A<1 thì A-1<0
=>\(\dfrac{x-3-x-5}{x+5}< 0\)
=>x+5>0
=>x>-5
c: Để A=(2x-3)/(x+1) thì \(\dfrac{2x-3}{x+1}=\dfrac{x-3}{x+5}\)
=>2x^2+10x-3x-15=x^2-2x-3
=>2x^2+7x-15-x^2+2x+3=0
=>x^2+9x-12=0
hay \(x=\dfrac{-9\pm\sqrt{129}}{2}\)
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 \(P=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(B=\left(\dfrac{2x+1}{x-1}+\dfrac{8}{x^2-1}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x^2-1}{5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{2x^2+2x+x+1+8-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{2x^2+3x+9-x^2+2x-1}{5}\)
\(=\dfrac{x^2+5x+8}{5}\)
Ta có: \(x^2+5x+8\)
\(=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{7}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\)
Ta có: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\)
\(\Leftrightarrow x^2+5x+8>0\forall x\)
\(\Leftrightarrow\dfrac{x^2+5x+8}{5}>0\forall x\) thỏa mãn ĐKXĐ(đpcm)
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
\(a,A=\dfrac{x+1+2-2x+5-x}{\left(1-x\right)\left(x+1\right)}\cdot\dfrac{\left(1-x\right)\left(x+1\right)}{2x-1}\left(x\ne1;x\ne-1;x\ne\dfrac{1}{2}\right)\\ A=\dfrac{8-2x}{2x-1}\\ b,A>0\Leftrightarrow\dfrac{8-2x}{2x-1}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}8-2x>0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}8-2x< 0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 4\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x< 4\\x\in\varnothing\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 4\)
Bạn ghi lại câu b cho mình đc ko, câu b bị mất 1 đoạn ở dưới rồi