Bài 1.
a) x3 + 2x2 - 3x - 6 = ( x3 + 2x2 ) - ( 3x + 6 ) = x2( x + 2 ) - 3( x + 2 ) = ( x + 2 )( x2 - 3 )
b) ( x - 9 )( x - 7 ) + 1 = x2 - 16x + 63 + 1 = x2 - 16x + 64 = ( x - 8 )2
c) ( x2 + x - 1 )2 + 4x2 + 4x
= ( x2 + x - 1 )2 + 4( x2 + x ) (1)
Đặt t = x2 + x
(1) <=> ( t - 1 )2 + 4t
= t2 - 2t + 1 + 4t
= t2 + 2t + 1
= ( t + 1 )2
= ( x2 + x + 1 )2
d) ( x2 + y2 - 17 )2 - 4( xy - 4 )2
= ( x2 + y2 - 17 )2 - 22( xy - 4 )2
= ( x2 + y2 - 17 )2 - [ 2( xy - 4 ) ]2
= ( x2 + y2 - 17 )2 - ( 2xy - 8 )2
= [ ( x2 + y2 - 17 ) - ( 2xy - 8 ) ][ ( x2 + y2 - 17 ) + ( 2xy - 8 ) ]
= ( x2 + y2 - 17 - 2xy + 8 )( x2 + y2 - 17 + 2xy - 8 )
= [ ( x2 - 2xy + y2 ) - 17 + 8 ][ ( x2 + 2xy + y2 ) - 17 - 8 ]
= [ ( x - y )2 - 9 ][ ( x + y )2 - 25 ]
= [ ( x - y )2 - 32 ][ ( x + y )2 - 52 ]
= ( x - y - 3 )( x - y + 3 )( x + y - 5 )( x + y + 5 )
Bài 2.
ĐK : x, y ∈ Z
a) x + 2y = xy + 2
<=> x + 2y - xy - 2 = 0
<=> ( x - xy ) - ( 2 - 2y ) = 0
<=> x( 1 - y ) - 2( 1 - y ) = 0
<=> ( 1 - y )( x - 2 ) = 0
+) Nếu 1 - y = 0 => y = 1 và nghiệm đúng với mọi x ∈ Z
+) Nếu x - 2 = 0 => x = 2 và nghiệm đúng với mọi y ∈ Z
Vậy phương trình có hai nghiệm
1. \(\hept{\begin{cases}y=1\\\forall x\inℤ\end{cases}}\); 2. \(\hept{\begin{cases}x=2\\\forall y\inℤ\end{cases}}\)
b) xy = x + y
<=> xy - x - y = 0
<=> ( xy - x ) - ( y - 1 ) - 1 = 0
<=> x( y - 1 ) - ( y - 1 ) = 1
<=> ( y - 1 )( x - 1 ) = 1
Ta có bảng sau :
| y-1 | 1 | -1 |
| x-1 | 1 | -1 |
| y | 2 | 0 |
| x | 2 | 0 |
Các nghiệm trên đều thỏa mãn ĐK
Vậy ( x ; y ) = { ( 2 ; 2 ) , ( 0 ; 0 ) }
a,\(x^3+2x^2-3x-6\)
\(=\left(x^3+2x^2\right)-\left(3x+6\right)\)
\(=x^2\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-3\right)\)
b,\(\left(x-9\right)\left(x-7\right)+1\)
\(=x^2-7x-9x+63+1\)
\(=x^2-16x+64\)
\(=\left(x-8\right)^2\)
c,\(\left(x^2+x-1\right)^2+4x^2+4x\)
\(=\left(x^2+x-1\right)^2+4\left(x^2+x\right)\)(*)
Đặt \(a=x^2+x\)ta đc:
(*)=\(\left(a-1\right)^2+4a\)
\(=a^2-2a+1+4a\)
\(=a^2+2a+1\)
\(=\left(a+1\right)^2\)
\(=\left(x^2+x+1\right)^2\)
d,\(\left(x^2+y^2-17\right)^2-4\left(xy-4\right)^2\)
\(=\left(x^2+y^2-17\right)^2-\left[2\left(xy-4\right)\right]^2\)
\(=\left(x^2+y^2-17\right)^2-\left(2xy-8\right)^2\)
\(=\left(x^2+y^2-17-2xy+8\right)\left(x^2+y^2-17+2xy-8\right)\)
\(=\left(x^2-2xy+y^2-9\right)\left(x^2+2xy+y^2-25\right)\)
\(=\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-25\right]\)
\(=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-5\right)\left(x+y+5\right)\)
