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a: \(P=\left(\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)\cdot\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\cdot\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{1}{x+2}\cdot\dfrac{x^3-x-2x+2}{x^2+x+1}\right)\)
\(=\left(\dfrac{x}{x+2}-\dfrac{x^2-2x+4}{\left(x+2\right)^2}\right):\left(\dfrac{1}{x+2}\cdot\dfrac{x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)}{x^2+x+1}\right)\)
\(=\dfrac{x^2+2x-x^2+2x-4}{\left(x+2\right)^2}:\left(\dfrac{1}{x+2}\cdot\dfrac{\left(x-1\right)\left(x^2+x-2\right)}{x^2+x+1}\right)\)
\(=\dfrac{4x-4}{\left(x+2\right)^2}:\left(\dfrac{1}{x+2}\cdot\dfrac{\left(x-1\right)\left(x+2\right)\left(x-1\right)}{x^2+x+1}\right)\)
\(=\dfrac{4\left(x-1\right)}{\left(x+2\right)^2}\cdot\dfrac{x^2+x+1}{\left(x-1\right)^2}=\dfrac{4\left(x^2+x+1\right)}{\left(x+2\right)^2\left(x-1\right)}\)
b: Để P>0 thì x-1>0
hay x>1
\(B=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{x^2-3x}{2x^2-x^3}\left(ĐKXĐ:x\ne2;-2;0\right)\)
a)\(B=\left(-\frac{\left(x+2\right)^2}{x^2-4}-\frac{4x^2}{x^2-4}+\frac{\left(x-2\right)^2}{x^2-4}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(B=\left(\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{x^2-4}\right).\frac{-x\left(x-2\right)}{\left(x-3\right)}\)
\(B=\left(\frac{-x^2-4x-4-4x^2+x-4x+4}{\left(x-2\right)\left(x+2\right)}\right).-\frac{x\left(x-2\right)}{x-3}\)
\(B=\frac{-5x^2-7x}{\left(x+2\right)}.\frac{-x}{x-3}\)
\(B=\frac{\left(-5x^2-7x\right)-x}{\left(x+2\right)\left(x-3\right)}\)
\(B=\frac{5x^3+7x^2}{\left(x+2\right)\left(x+3\right)}\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\) ĐKXD: \(x\ne\pm2,x\ne0,x\ne3\)
\(\Leftrightarrow\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{2-x}{2+x}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)
\(\Leftrightarrow\left(\frac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right):\left(\frac{x-3}{x\left(2-x\right)}\right)\)
\(\Leftrightarrow\left(\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\right)\cdot\left(\frac{x\left(2-x\right)}{x-3}\right)\)
\(\Leftrightarrow\frac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2-x\right)}{x-3}\)
\(\Leftrightarrow\frac{4x^2}{x-3}\)
b, Để A>0 thì \(\frac{4x^2}{x-3}>0\)
\(\Rightarrow4x^2>0\)
\(\Rightarrow x>0\)
c, Ta có
\(\left|x-7\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\left(l\right)\end{cases}}}\)
Với \(x=11\Rightarrow\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)
a, ĐKXĐ : \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)hoặc \(2x^2-x^3\ne0\)hay \(x\ne\pm2;0\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(=\left(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)
\(=\frac{-x^2-2x-1-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}:\frac{x-3}{x\left(2-x\right)}\)
\(=\frac{-4x^2-6x+3}{\left(x-2\right)\left(x+2\right)}.\frac{-x\left(x-2\right)}{x-3}=\frac{\left(-4x^2-6x+3\right)\left(-x\right)}{\left(x+2\right)\left(x-3\right)}=\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}\)
b, Ta có : A > 0 hay \(\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}>0\)
\(\Leftrightarrow x\left(4x^2+6x-3\right)>0\)
\(\Leftrightarrow4x^2+6x-3>0\) bạn xem lại bài mình có chỗ nào sai ko nhé !!!
c, Ta có : \(\left|x-7\right|=4\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)
TH1 : Thay x = 11 vào phân thức trên : ...
TH2 : Thay x = 3 vào phân thức trên : .... tự làm
a. Ta có :
\(x^4-x^3-2x-4\)
\(=x^4-2x^3+x^3-2x-4\)
\(=x^3\left(x-2\right)+\left(x^3-2x^2\right)+\left(x^2-4\right)+\left(x^2-2x\right)\)
\(=x^3\left(x-2\right)+x^2\left(x-2\right)+\left(x+2\right)\left(x-2\right)+x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3+x^2+x+2+x\right)\)
\(=\left(x-2\right)\left[\left(x^3+2x\right)+\left(x^2+2\right)\right]\)
\(=\left(x-2\right)\left[x\left(x^2+2\right)+\left(x^2+2\right)\right]\)
\(=\left(x-2\right)\left(x^2+2\right)\left(x+1\right)\)
Ta lại có :
\(2x^4-3x^3+2x^2-6x-4\) ... biến đổi tương tự ta được \(\left(x^2+2\right)\left(x-2\right)\left(2x+1\right)\)
Do đó với \(x\ne2;x\ne\frac{1}{2}\) thì \(P=\frac{\left(x^2+2\right)\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x^2+2\right)\left(2x+1\right)}=\frac{x+1}{2x+1}\) ( = 1/2 )
Cảm ơn Let Hate Him nha! Nhưng bạn có thể biến đổi nốt phần sau giúp mình được không?