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m . n 3 – 1 + m – n 3 = ( m n 3 – n 3 ) + ( m - 1 ) = n 3 ( m – 1 ) + ( m – 1 ) = n 3 + 1 m - 1 = ( n + 1 ) ( n 2 – n + 1 ) ( m – 1 )
Đáp án cần chọn là: A
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4 x 2 + 4 x – y 2 + 1 = ( ( 2 x ) 2 + 2 . 2 x + 1 ) – y 2
= ( 2 x + 1 ) 2 – y 2
= (2x + 1 – y)(2x + 1 + y)
= (2x – y + 1)(2x + y + 1)
Vậy đa thức trong chỗ trống là 2x – y + 1
Đáp án cần chọn là: B
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Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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\(A=n^4-n^3-6n^2+7n-21\)
\(A=n^4-3n^3+2n^3-6n^2+7n-21\)
\(A=n^3\left(n-3\right)+2n^2\left(n-3\right)+7\left(n-3\right)\)
\(A=\left(n^3+2n^2+7\right)\left(n-3\right)\)
\(A=n^4-n^3-6n^2+7n-21\)
\(A=n^4-3n^3+2n^3-6n^2+7n-21\)
\(A=n^3\left(n-3\right)+2n^2\left(n-3\right)+7\left(n-3\right)\)
\(A=\left(n^3+2n^2+7\right)\left(n-3\right)\)
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1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
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\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)
\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)
\(=3\left(a-b+c\right)\left(x+6y\right)^2\)
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\(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
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Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
n3 -n2 -7n +1=n2(n-1) - 7(n-1)-6=0 <=>(n-1).(n2-7)=6