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a \(x\cdot\left(a-2\right)-a\left(a-2\right)\)
\(\left(x-a\right)\cdot\left(a-2\right)\)
b ,\(x\left(x+1\right)-a\left(x+1\right)\)
\(\left(x-a\right)\cdot\left(x+1\right)\)
c ,\(2x\left(x+2a\right)+x+2a\)
(2x+1)(x+2a)
d, 2xy+x^2-ax-2ay
x(2y+x)-a(x+2y)
(x-a)(x+2y)
e, x^2(x+a)+x+a
(x^2+1)(x+a)
f, y^2(y+x^2)+z(x^2+y)
(y^2+z)(y+x^2)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( x2 - 25 )2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) ]2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) - ( x - 5 ) ][ ( x - 5 )( x + 5 ) + ( x - 5 ) ]
= ( x - 5 )( x + 5 - 1 )( x - 5 )( x + 5 + 1 )
= ( x - 5 )2( x + 4 )( x + 6 )
b) ( 4x2 - 25 )2 - 9( 2x - 5 )2
= ( 4x2 - 25 )2 - 32( 2x - 5 )2
= ( 4x2 - 25 )2 - ( 6x - 15 )2
= [ ( 4x2 - 25 ) - ( 6x - 15 ) ][ ( 4x2 - 25 ) + ( 6x - 15 ) ]
= ( 4x2 - 25 - 6x + 15 )( 4x2 - 25 + 6x - 15 )
= ( 4x2 - 6x - 10 )( 4x2 + 6x - 40 )
= ( 4x2 + 4x - 10x - 10 )( 4x2 + 16x - 10x - 40 )
= [ 4x( x + 1 ) - 10( x + 1 ) ][ 4x( x + 4 ) - 10( x + 4 ) ]
= ( x + 1 )( 4x - 10 )( x + 4 )( 4x - 10 )
= ( 4x - 10 )2( x + 1 )( x + 4 )
c) 4( 2x - 3 )2 - 9( 4x2 - 9 )2
= 22( 2x - 3 )2 - 32( 4x2 - 9 )2
= ( 4x - 6 )2 - ( 12x2 - 27 )2
= [ ( 4x - 6 ) - ( 12x2 - 27 ) ][ ( 4x - 6 ) + ( 12x2 - 27 ) ]
= ( 4x - 6 - 12x2 + 27 )( 4x - 6 + 12x2 - 27 )
= ( -12x2 + 4x + 21 )( 12x2 + 4x - 33 )
= ( -12x2 + 18x - 14x + 21 )( 12x2 - 18x + 22x - 33 )
= [ -12x( x - 3/2 ) - 14( x - 3/2 ) ][ 12x( x - 3/2 ) + 22( x - 3/2 ) ]
= ( x - 3/2 )( -12x - 14 )( x - 3/2 )( 12x + 22 )
= ( x - 3/2 )2( -12x - 14 )( 12x + 22 )
d) a6 - a4 + 2a3 + 2a2
= a2( a4 - a2 + 2a + 2 )
= a2( a4 - 2a3 + 2a3 + 2a2 - 4a2 + a2 + 4a - 2a + 2 )
= a2[ ( a4 - 2a3 + 2a2 ) + ( 2a3 - 4a2 + 4a ) + ( a2 - 2a + 2 ) ]
= a2[ a2( a2 - 2a + 2 ) + 2a( a2 - 2a + 2 ) + 1( a2 - 2a + 2 ) ]
= a2( a2 + 2a + 1 )( a2 - 2a + 2 )
= a2( a + 1 )2( a2 - 2a + 2 )
e) ( 3x2 + 3x + 2 )2 - ( 3x2 + 3x - 2 )2
= [ ( 3x2 + 3x + 2 ) - ( 3x2 + 3x - 2 ) ][ ( 3x2 + 3x + 2 ) + ( 3x2 + 3x - 2 ) ]
= ( 3x2 + 3x + 2 - 3x2 - 3x + 2 )( 3x2 + 3x + 2 + 3x2 + 3x - 2 )
= 4( 6x2 + 6x )
= 4.6x( x + 1 )
= 24( x + 1 )
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Bài 1
\(x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)+\left(-x^3-x^2-x\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
Bài 2
Ta có: \(\left(ax+b\right)\left(x^2+cx+1\right)=ax^3+bx^2+acx^2+bcx+ax+b\)
\(=ax^3+\left(b+ac\right)x^2+\left(bc+a\right)x+b=x^3-3x-2\)
\(\Rightarrow a=1\)
\(\Rightarrow b+ac=0\)
\(\Rightarrow bc+a=-3\)
\(\Rightarrow b=-2\)
Thay giá trị của \(a=1;b=-2\)vào \(b+ac=0\)ta được
\(\Leftrightarrow-2+c=0\Rightarrow c=2\)
Vậy \(a=1;b=-2;c=2\)
Bài 3
Ta có \(\left(x^4-3x^3+2x^2-5x\right)\div\left(x^2-3x+1\right)=x^2+1\left(dư-2x+1\right)\)
\(\Rightarrow b=2x-1\)
Bài 4 (cũng làm tương tự như bài 3 nhé )
Bài 5(bài nãy dễ nên bạn tự làm đi nhé)
Bài 6
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)\(\Rightarrow a-b=0\Rightarrow a=b\)
Bài 7
\(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a-b=0\Rightarrow a=b\)
\(\Rightarrow b-c=0\Rightarrow b=c\)
\(\Rightarrow a-c=0\Rightarrow a=c\)
Vậy \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(ax-2x-a^2+2a\)
\(=x\left(a-2\right)-a\left(a-2\right)\)
\(=\left(x-a\right)\left(a-2\right)\)
Vậy....
b/ \(x^2+x-ax-a\)
\(=x\left(x+1\right)-a\left(x+1\right)\)
\(=\left(x+1\right)\left(x-a\right)\)
Vậy...
d/ \(2xy-ax+x^2-2ay\)
\(=2y\left(x-a\right)+x\left(x-a\right)\)
\(=\left(x-a\right)\left(2y+x\right)\)
e/ \(x^3+ax^2+x+a\)
\(=x^2\left(x+a\right)+\left(x+a\right)\)
\(=\left(x^2+1\right)\left(x+a\right)\)
Vậy...
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a) ax - 2x - a2 + 2a
= ( ax - 2x ) - ( a2 - 2a )
= x ( a - 2 ) - a ( a - 2 )
= ( a - 2 ) ( x - a )
b) x2 + x - ax - a
= ( x2 + x ) - ( ax + a )
= x ( x + 1 ) - a ( x + 1 )
= ( x + 1 ) ( x - a )
Hok Tốt!!!
a) ax -2x- a2+ 2a
= (ax -2x ) -(a2 -2a )
= x(a-2) -a ( a-2 )
= (x-a) (a-2)
b) x2 +x -ax -a
=( x2 +x ) - ( ax +a )
= x( x+1 ) -a ( x+1 )
= ( x-a ) (x+ 1)
c) 2x2 +4ax +x +2a
=( 2x2 + 4ax ) + ( x+ 2a )
= 2x ( x+ 2a ) + ( x+2a )
= ( 2x +1 ) (x+2a )
d) 2xy -ax +x2 - 2ay
= (2xy -2ay ) + ( -ax + x2 )
= 2y( x-a ) + x ( x-a)
= ( 2y +x ) ( x -a )
Đáp án
=a^6x^6-2a^3x^7+11a^2x^10
Học tốt
ý nhầm sửa hộ mik nha
=a^7x^6-2a^4x^7+11a^2x^10
hì hì bn thông cảm nha