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\(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\\ ---\\ 4x^2-4x-3\\ =4x^2-4x+1-4\\ =\left(2x-1\right)^2-2^2=\left(2x-1-2\right)\left(2x-1+2\right)\\ =\left(2x-3\right)\left(2x+1\right)\)
1: =(2x)^2-2*2x*1+1^2
=(2x-1)^2
2: =4x^2-6x+2x-3
=2x(2x-3)+(2x-3)
=(2x-3)(2x+1)
\(4x^4+4x^3+5x^2+2x+1\)
\(=4x^4+2x^3+2x^2+2x^3+x^2+2x^2+x+1\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
C1: \(4x^2-4x+1=\left(2x-1\right)^2\) (Hằng đẳng thức bạn ạ)
C2: \(4x^2-4x+1\)
=\(4x^2-2x-2x+1\)
=\(2x\left(2x-1\right)-\left(2x-1\right)\)
=\(\left(2x-1\right)\left(2x-1\right)\)
=\(\left(2x-1\right)^2\)
Mình chắc 100% đó . **** mình bạn na !!!
\(x^3+4x^2+4x+1\)
\(=x^3+3x^2+x+x^2+3x+1\)
\(=x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)\)
\(=\left(x+1\right)\left(x^2+3x+1\right)\)
= ( 4x^2 + 4x + 1 ) - y^2
= ( 2x + 1 )^2 - y^2
= ( 2x + 1 - y)( 2x + 1 + y)
1) ( 4x + 1 )2 + ( 4x - 1 )2 - 2( 4x + 1 ).( 4x - 1 )
= ( 4x + 1 - 4x - 1 )2
= 22
= 4
2) 4x2 - 9 + ( 2x + 3 )
= ( 2x )2 - 32 + ( 2x + 3 )
= ( 2x + 3 ).( 2x - 3 ) + ( 2x + 3 )
= ( 2x + 3 ). ( 2x - 3 + 1 )
= ( 2x + 3 ) .( 2x - 2 )
= 2.( 2x + 3 ) .( x - 1 )
1, (4x+1)^2 + (4x-1)^2 - 2(4x+1)(4x-1)
=[(4x+1)-(4x-1)]^2
=(4x+1-4x+1)^2
=2^2
=4
2, 4x^2 - 9 +(2x+3)
=(4x^2 - 9)+(2x+3)
=(2x+3)(2x-3)+(2x+3)
=(2x+3)(2x-3+1)
=(2x+3)(2x-2)
=2(x-1)(2x+3)
=.= hok tốt!!
\(\left(4x-1\right)^2-4x^2\)
\(=\left(4x-1\right)^2-\left(2x\right)^2\)
\(=\left(4x-1-2x\right)\left(4x-1+2x\right)\)
\(=\left(2x-1\right)\left(6x-1\right)\)
`(4x-1)^2-4x^2=(4x-1)^2-(2x)^2=(4x-1-2x)(4x-1+2x)=(2x-1)(6x-1)`