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\(f( x;y) = ax + 2x + ay + 2y + 4 = a^2\)
=> \(( a + 2 )( x + y ) = a^2 -4\)
=> \(( a + 2 )( x + y ) = ( a-2 )( a + 2 )\)
=> \(( a + 2 )( x + y ) - ( a-2 )( a + 2 )=0\)
=> \(( a + 2 )[ x + y - ( a-2 )] = 0\)
=> \(\left[\begin{matrix} x+y - ( a-2 )=0\\ a+2=0\end{matrix}\right.\)
=> \(\left[\begin{matrix} x+y = ( a-2 )\\ a=-2\end{matrix}\right.\)
Như vậy , nếu \(x+y=a-2\) thì \(f( x;y) = ax + 2x + ay + 2y + 4 = a^2\)
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\(ax+2x+ay+2y+4\)
\(\Leftrightarrow x\left(a+2\right)+y\left(a+2\right)+4\)
\(=\left(x+y\right)\left(a+2\right)+4\)
\(=a^2-4+4=a^2\)
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\(A=2x^4+4x^3y-x^3y-4x^3+x^2y^2-2x^2y-2x+2x+3\)
\(A=2.\left(x^4+2x^3y+x^2y^2\right)-x^2y^2-x^3y-4x^3-2x^2y+3\)
\(A=2.\left(x^2+xy\right)^2-\left(x^2y^2+x^3y\right)-\left(4x^3+2x^2y\right)+3\)
\(A=2.x^2.\left(x+y\right)^2-x^2y\left(y+x\right)-2x^2\left(2x+y\right)+3\)
\(A=8.x^2-2.x^2y-2x^2\left(x+2\right)+3=8x^2-2x^2\left(2-x\right)-2x^3-4x^2+3\)
\(A=8x^2-4x^2 +2x^3-2x^3-4x^2+3=3\)là hằng số
=> ĐPCM
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Bài 1:
a) \(f\left(x\right)=2x\left(x^2-3\right)-4\left(1-2x\right)+x^2\left(x-1\right)+\left(5x+3\right)\)
\(=2x^3-6x-4+8x+x^3-x^2+5x+3\)
\(=x^3-x^2+7x-1\)
\(g\left(x\right)=-3\left(1-x^2\right)-2\left(x^2-2x+1\right)\)
\(=-3+3x^2-2x^2+4x-2\)
\(=x^2+4x-5\)
b) \(h\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(=x^3-x^2+7x-1-x^2-4x+5\)
\(=x^3-2x^2+3x-4\)
ax+2x+ay+2y+4
=a(x+y)+2(x+y)+4
=(a+2)(x+y)+4
=(a+2)(a-2)+4
=a^2-4+4
=a^2
Với a - 2 = x + y ta có :
ax + 2x + ay + 2y + 4
= ( ax + ay ) + ( 2x + 2y ) + 4
= a ( x + y ) + 2 ( x + y ) + 4
= ( a + 2 ) ( x + y ) + 4
= ( a + 2 ) ( a - 2 ) + 4
= a2 - 4 + 4 = a2 (đpcm)