Phân tích đa thức thành nhân tử PP dùng hằng đẳng thức;
1) (3x-1)2- 4
2) (x+y)2- x2
3) 100- (2x- y)2
4) (2x-1)2 - (x- 1)2
5) 4(x+ 6)2- 9(1+ x)2
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\(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=\left(x+2\right)3x\)
TL:
\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)\left(2x+1-x+1\right)\)
\(=3x.\left(x+2\right)\)
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) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
Bài 2:
7) \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
8) \(x^3+64=\left(x+4\right)\left(x^2+4x+16\right)\)
9) \(\left(a+b\right)^2-\left(2a-b\right)^2=\left(a+b+2a-b\right)\left(a+b-2a+b\right)=3a\left(-a+2b\right)\)
10) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a\cdot2b=4ab\)
11) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2a\left(3a^2+b^2\right)\)
12) \(\left(6x-1\right)^2-\left(3x+2\right)^2=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)=\left(9x+1\right)\left(3x-3\right)\)
\(10x-25-x^2=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2.x.5+5^2\right)=-\left(x-5\right)^2\)
\(\Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
8x3- 125= (2x)3- 53= (2x-5)[(2x)2+2x5+52 ]=(2x-5)(4x2+10x+25)
\(2x^2-7x+3\)
\(=2\left(x^2-\frac{7}{2}x+\frac{3}{2}\right)\)
Vậy thôi đâu cần dùng HĐT
a) ( 3x - 1 )2 - 4
= ( 3x - 1 ) - 22
= ( 3x - 1 - 2 )( 3x - 1 + 2 )
= ( 3x - 3 )( 3x + 1 )
= 3( x - 1 )( 3x + 1 )
b) ( x + y )2 - x2
= ( x + y - x )( x + y + x )
= y( 2x + y )
c) 100 - ( 2x - y )2
= 102 - ( 2x - y )2
= [ 10 - ( 2x - y ) ][ 10 + ( 2x - y ) ]
= ( 10 - 2x + y )( 10 + 2x - y )
d) ( 2x - 1 )2 - ( x - 1 )2
= [ ( 2x - 1 ) - ( x - 1 ) ][ ( 2x - 1 ) + ( x - 1 ) ]
= ( 2x - 1 - x + 1 )( 2x - 1 + x - 1 )
= x( 3x - 2 )
e) 4( x + 6 )2 - 9( 1 + x )2
= 22( x + 6 )2 - 32( 1 + x )2
= ( 2x + 12 )2 - ( 3 + 3x )2
= [ ( 2x + 12 ) - ( 3 + 3x ) ][ ( 2x + 12 + ( 3 + 3x ) ]
= ( 2x + 12 - 3 - 3x )( 2x + 12 + 3 + 3x )
= ( 9 - x )( 5x + 15 )
= 5( 9 - x )( x + 3 )