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A=x2y−y+xy2−xx2y−y+xy2−x
A=(x2y−y)+(xy2−x)(x2y−y)+(xy2−x)
A=y(x2−1)+x(y2−1)y(x2−1)+x(y2−1)
A=y(x-1)(x+1)+x(y-1)(y+1)
thay x=-5 và y=2 ta có:
A=2(-5-1)(-5+1) - 5(2-1)(2+1)
A=2 . (-6) . (-4) - 5 . 3
A=48 - 15
A= 33
\(x^2.y-y+x.y^2-x=\left(-5\right)^2.2-2+\left(-5\right).2^2-\left(-5\right)\)
\(=25.2-2-5.4+5=50-2-20+5=33\)
1.a) xy + 2y - x2 + 4
= y ( x + 2 ) - ( x2 - 4 ) = y ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 )( y - x + 2 )
b) 2x2 + y2 + 3xy
= ( 2x2 + 2xy ) + ( y2 + xy )
= 2x ( x + y ) + y ( x + y )
= ( x + y ) ( 2x + y )
2.
x - y = 5 \(\Rightarrow\)( x - y )2 = 25 \(\Rightarrow\)x2 + y2 = 25 + 2xy = 25 + 2.3 = 31
A = ( x + y )2 = x2 + y2 + 2xy = 31 + 6 = 37
a) Ta có:
VT = (x - y)² + 4xy
= x² - 2xy + y² + 4xy
= x² + 2xy + y²
= (x + y)²
= VP
b) Ta có:
(x + y)² = (x - y)² + 4xy
= 5² + 4.3
= 25 + 12
= 37
\(a,N=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ N=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\\ b,N=\left(x+y\right)^2-2xy=0-2\cdot1=-2\)
ĐKXĐ: \(x\ne y\)
a) \(N=\dfrac{x^2+y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}:\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=x^2+y^2\)
b) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)
\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)
$(x+y)^2\\=x^2+2xy+y^2\\=(x^2-2xy+y^2)+4xy\\=(x-y)^2+4xy\\=5^2+4.3\\=25+12\\=37$
`A=(x+y)^2=x^2+2xy+y^2=(x^2-2xy+y^2)+4xy=(x-y)^2+4xy`
Thay `x-y=5;xy=3` được: `A=5^2+4.3=37`