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a) 16x2 - 9
= ( 4x )2 - 32
= ( 4x - 3 )( 4x + 3 )
b) 9a2 - 25b4
= ( 3a )2 - ( 5b2 )2
= ( 3a - 5b2 )( 3a + 5b2 )
c) 81 - y4
= 92 - ( y2 )2
= ( 9 - y2 )( 9 + y2 )
= ( 32 - y2 )( 9 + y2 )
= ( 3 - y )( 3 + y )( 9 + y2 )
d) ( 2x + y )2 - 1
= ( 2x + y )2 - 12
= ( 2x + y - 1 )( 2x + y + 1 )
e) ( x + y + z )2 - ( x - y - z )2
= [ x + y + z - ( x - y - z ) ][ x + y + z + ( x - y - z ) ]
= [ x + y + z - x + y + z ][ x + y + z + x - y - z ]
= [ 2y + 2z ].2x
= 2[ y + z ].2x
= 4x[ y + z ]
a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)
b) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-3\right)\left(x-1\right)\)
c) \(x^2+5x+4=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+4\right)\left(x+1\right)\)
d) \(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(b,x^2+4x+3=x^2+3x+x+3.\)
\(=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)
\(c,16x-5x^2-3=x-5x^2+15x-3\)
\(=x\left(1-5x\right)+3\left(5x-1\right)\)
\(=\left(x+3\right)\left(1-5x\right)\)
\(d,x^4+4=x^4+4x^2+4-4x^2=\left(x+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
a) x5 + x + 1
= x5 - x4 + x4 - x3 + x3 - x2 + x2 + x + 1
= ( x5 + x4 + x3) - ( x4 + x3 + x2) + (x2 + x +1)
= x3( x2 + x + 1) - x2( x2 + x + 1) + (x2 + x +1)
= ( x2 + x +1).( x3 - x2 + 1)
b) ( x2 + x)2 -2(x2 + x) -15
=( x2 + x)2 -2(x2 + x).1 + 1- 16
=( x2 + x - 1)2 - 42
=( x2 + x - 1 - 4).( x2 + x - 1+ 4)
=(x2 + x - 5).( x2 + x + 3)
c) x4 + 5x3 + 10x - 4
= (x2)2 - 22 + 5x.( x2 + 2)
=( x2 -2).(x2 + 2) + 5x.( x2 + 2)
= ( x2 + 2).(x2 -2 + 5x)
d) x8 + x7 + 1
= x8 + x7 + x6 - x6 + 1
= x6 ( x2 + x + 1) - ( x6 - 1)
= x6( x2 + x + 1) - ( x3 - 1).(x3 + 1)
= x6( x2 + x + 1) - ( x- 1).( x2 + x + 1).(x3 + 1)
= ( x2 + x + 1).[ x6 -( x- 1).(x3 + 1)]
= ( x2 + x + 1).( x6 - x4 + x3 - x +1)
cau b)
\(B=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-16=\left(x^2+x-1\right)^2-4^2\)\(B=\left(x^2+x-3\right)\left(x^2+x+1\right)\)
\(B=\left[\left(x+\dfrac{1}{2}\right)^2-\left(\sqrt{\dfrac{13}{4}}\right)^2\right]\left(x^2+x+1\right)\)
\(B=\left(x+\dfrac{1-\sqrt{13}}{2}\right)\left(x+\dfrac{1+\sqrt{13}}{2}\right)\left(x^2+x+1\right)\)
Sai r pn...kquả là 3 nhưg mik k pit cách lm
ko có trả lời à