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`(x+1)(x^2-x+1)-3x(x-2)`
`= x(x^2-x+1)+1(x^2-x+1) -[3x*x+3x*(-2)]`
`= x*x^2+x*(-x)+x*1+x^2-x+1-(3x^2-6x)`
`=x^3-x^2+x+x^2-x+1-3x^2+6x`
`= x^3-3x^2+6x+1`
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a) \(|x|-x\)
\(\Rightarrow\orbr{\begin{cases}x< 0\rightarrow\left|x\right|-x=2\left|x\right|\\x>0\rightarrow\left|x\right|-x=0\end{cases}}\)
\(\Rightarrow x=0\rightarrow x=0\)
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\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right).x^2.\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
Ta có:
\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)x^2\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left[\left(2x^3-x^2+1\right)+\left(x^2-2x^3\right)\right]\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
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(2x - 1)2 + 2(2x - 1)(x + 1) + (x - 1)2 (Dễ dàng nhận thấy đây là HĐT số 1)
= (2x -1 + x - 1)2
= (3x - 2)2
Bài làm
x( 1 - x ) + ( x + 1 )( x - 2 )
= x - x2 + x2 - 2x + x - 2
= ( x + x - 2x ) + ( -x2 + x ) - 2
= -2
Vậy giá trị của đa thức trên là ( -2 )
# Học tốt #