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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
Ta có \(x-y=5\Rightarrow x^2-2xy+y^2=25\Rightarrow x^2+y^2=25+2xy=25+2.3=31\)
\(\left(x+y\right)^2=\left(x^2+y^2\right)+2xy=31+2.3=37\)
\(a,N=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ N=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\\ b,N=\left(x+y\right)^2-2xy=0-2\cdot1=-2\)
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) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)
\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)
a)Ta có:\(x-y=2\Rightarrow\left(x-y\right)^2=4\Rightarrow\left(x^2+y^2\right)-2xy=4\Rightarrow4-2xy=4\Rightarrow2xy=0\Rightarrow xy=0\)
Khi đó ta có:\(x^5y=xy^5=xy\left(x^4-y^4\right)=0\)
x2 - 5x - 2xy + 5y + y2 + 4
= (x2 - 2xy + y2) - (5x - 5y) + 4
= (x2 - xy - xy + y2) - 5.(x - y) + 4
= (x - y)2 - 5.1 + 4
= 1 - 5 + 4
= 0
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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
$(x+y)^2\\=x^2+2xy+y^2\\=(x^2-2xy+y^2)+4xy\\=(x-y)^2+4xy\\=5^2+4.3\\=25+12\\=37$
`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`