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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
2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)
Khi |x - 1| = 2
=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
Khi x = - 1 (không thỏa mãn) => Không tìm được A
b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
Đẻ P < 8
=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)
=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)
Vậy x > - 1 thì P < 8
\(B=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{x^2-3x}{2x^2-x^3}\left(ĐKXĐ:x\ne2;-2;0\right)\)
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. 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?