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a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)
=\(x^3\left(x-1\right)+x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)
=\(\left(x-1\right)\left(x^3+x^2-x+1\right)\)
b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)
=\(a^2b^2-2ab+1+a^2+2ab+b^2\)
=\(a^2b^2+a^2+b^2+1\)
=\(a^2\left(b^2+1\right)+\left(b^2+1\right)\)
=\(\left(b^2+1\right)\left(a^2+1\right)\)
c,\(x^4+2x^3+2x^2+2x+1\)
=\(x^4+x^3+x^3+x^2+x^2+x+x+1\)
=\(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
=\(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
=\(\left(x+1\right)^2\left(x^2+1\right)\)
mk ghi đáp án, còn lại bạn tự biến đổi
a) \(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4=\left(x+1\right)\left(x+2\right)^2\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
d) \(4x^4+1=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
e) \(x^4-7x^3+14x^2-7x+1=\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)
mk làm chi tiết theo yêu của của người hỏi đề:
a) \(2x^3-x^2+5x+3\)
\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4\)
\(=\left(x^3+4x^2+4x\right)+\left(x^2+4x+4\right)\)
\(=x\left(x^2+4x+4\right)+\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
1. \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)
\(=-7x-13\)
2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)
\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)
3. \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2.\left(2x-1\right)+\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)
\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
Bài 1.
B = ( x - 2 )( x + 2 )( x + 3 ) - ( x + 1 )3
= ( x2 - 4 )( x + 3 ) - ( x3 + 3x2 + 3x + 1 )
= x3 + 3x2 - 4x - 12 - x3 - 3x2 - 3x - 1
= -7x - 13
Bài 2.
64 - x2 - y2 + 2xy
= 64 - ( x2 - 2xy + y2 )
= 82 - ( x - y )2
= ( 8 - x + y )( 8 + x - y )
Bài 3.
2x3 - x2 + 2x - 1 = 0
<=> ( 2x3 - x2 ) + ( 2x - 1 ) = 0
<=> x2( 2x - 1 ) + 1( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 1 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( vì x2 + 1 ≥ 1 > 0 ∀ x )
1 ) \(a\left(m+n\right)+b\left(m+n\right)\)
\(=\left(a+b\right)\left(m+n\right)\)
2 ) \(a^2\left(x+y\right)-b^2\left(x+y\right)\)
\(=\left(a^2-b^2\right)\left(x+y\right)\)
\(=\left[\left(a-b\right).\left(a+3\right)\right]\left(x+y\right)\)
3 ) \(6a^2-3a+12ab\)
\(=3a.2a-3a+3a.4b\)
\(=3a.\left(2a-1+4b\right)\)
4 ) \(2x^2y^4-2x^4y^2+6x^3y^3\)
\(=2x^2y^2.y^2-2x^2y^2.x^2+2x^2y^2.3xy\)
\(=2x^2y^2\left(y^2-x^2+3xy\right)\)
5 ) \(\left(x+y\right)^3-x\left(x+y\right)^2\)
\(=\left(x+y\right)^2.\left(x+y-x\right)\)
\(=\left(x+y\right)^2.y\)
1)a(m+n)+b(m+n)
=(a+b)(m+n)
2)a2(x+y)-b2(x+y)
=(a2-b2)(x+y)
3)6a2-3a+12ab
=3a.2a-3a.(1-4b)
=3a.(2a-1+4b)
5)(x+y)3-x(x+y)2
=(x+y)(x+y)2-x(x+y)2
=(x+y)2(x+y-x)
a) \(x^3-\frac{1}{4}x=x\left(x^2-\frac{1}{4}\right)=x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)
b) \(\left(2x-1\right)^2-\left(x+3\right)^2=\left(2x-1-x-3\right)\left(2x-1+x+3\right)=\left(x-4\right)\left(3x+2\right)\)