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Bài 1 :
a) \(x\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x+1\right)\)
\(=\left(x^3-x\right)-\left(x^3+x^2-x-1\right)\)
\(=x^3-x-x^3-x^2+x+1\)
\(=1-x^2\)
b) \(\left(x+1\right)\left(x-2\right)-\left(2x-1\right)\left(x+2\right)+2x\left(x-1\right)\)
\(=\left(x^2-x+2\right)-\left(2x^2+3x-2\right)+\left(2x^2-2x\right)\)
\(=x^2-x+2-2x^3-3x+2+2x^3+2x\)
\(=x^2-2x+4\)
\(=\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\)
c) \(\left(x^2+2x-1\right)\left(x+2\right)-\left(x-1\right)\left(2x+1\right)\)
\(=\left(x^3+4x^2+3x-2\right)-\left(2x^2-x-1\right)\)
\(=x^3+4x^2+3x-2-2x^3+x+1\)
\(=-x^3+4x^2+4x-1\)
Bài 1
\(a)x\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x+1\right)\\ =\left(x+1\right)\left[x\left(x-1\right)-\left(x^2-1\right)\right]\\ =\left(1+x\right)\left(x^2-x-x^2+1\right)\\ =\left(1+x\right)\left(1-x\right)\\ =1-x^2\)
\(b)\left(x+1\right)\left(x-2\right)-\left(2x-1\right)\left(x+2\right)+2x\left(x-1\right)\\ =x^2-2x+x-2-\left(2x^2+4x-x-2\right)+2x^2-2x\\ =x^2-2x+x-2-(2x^2+3x-2)+2x^2-2x\\ =x^2-2x+x-2-2x^2-3x+2+2x^2-2x\\ =x^2-6x\)
\(c)\left(x^2+2x-1\right)\left(x+2\right)-\left(x-1\right)\left(2x+1\right)\\ =x^3+2x^2+2x^2+4x-x-2-\left(2x^2+x-2x-1\right)\\ =x^3+2x^2+2x^2+4x-x-2-\left(2x^2-x-1\right)\\ =x^3+2x^2+2x^2+4x-x-2-2x^2+x+1\\ =x^3+2x^2+4x-1\)
\(2x^4-x^3-2x^2-x+2=0\)
\(\Leftrightarrow2x^4-4x^3+2x^2+3x^3-6x^2+3x-4+2x^2-4x+2=0\)
\(\Leftrightarrow2x^2\left(x^2-2x+1\right)+3x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+3x+2\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x=2=0\left(vn\right)\\x^2-2x+1=0\Rightarrow x=1\end{matrix}\right.\)
Bạn tự thay \(x=1\) vào tính A