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Ta có:(x4+y4)=(x2+y2)2-2.x2.y2
=(x2+y2)2-2.xy.xy
=152-2.6.6
=225-72
=153
x4+y4
=[(x2)2+2x2y2+(y2)2]-2x2y2
=(x2+y2)-2x2y2
=(x2+y2)-2xy.xy
Ma x2+y2=15 va xy=6
=152-2.6.6
=225-72
=153
Nho k nha
Ta có: x.y = 15
=> x = \(\frac{15}{y}\)
Ta có x + y = -8
\(\frac{15}{y}\)+ y= 8
=> 15 + \(y^2\)= 8y => \(y^2-8y+15=0\)
=> y = 3 hoặc y = 5
=> y = 3 => x=5
y=5 => x=3
\(x^2+y^2=3^2+5^2=34\)
\(x^2+y^2=x^2+2xy+y^2=\left(x+y\right)^2-2xy\)
Vì x+y=-8,xy=15 nên:
\(\left(x+y\right)^2+2xy=\left(-8\right)^2+2.15=34\)
\(x^2y+xy^2+x+y=2018\)
\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)=2018\)
\(\Leftrightarrow\left(xy+1\right)\left(x+y\right)=2018\Leftrightarrow12\left(x+y\right)=2018\)
\(\Leftrightarrow x+y=\frac{1009}{6}\)
\(x^2+y^2=\left(x+y\right)^2-2xy=\left(\frac{1009}{6}\right)^2-2.11=...\)
Trả lời :
Ta có :
\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
Hok tốt
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y+2\right)\left(x+y+5\right).\)
b) \(x^2y+xy^2+x+y=2010\)
\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)=2010\)
\(\Leftrightarrow11\left(x+y\right)+1\left(x+y\right)=2010\)
\(\Leftrightarrow12\left(x+y\right)=2010\)
\(\Leftrightarrow x+y=\frac{335}{2}\)
\(\Leftrightarrow\left(x+y\right)^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+2xy+y^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2+22=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2=\frac{112137}{4}.\)
Vậy \(x^2+y^2=\frac{112137}{4}.\)
a,\(x^2+2xy+7x+7y+y^2+10=\left(x^2+2xy+y^2\right)+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
b,\(x^2y+xy^2+x+y=2010\Rightarrow xy\left(x+y\right)+x+y=2010\)
\(\Rightarrow12\left(x+y\right)=2010\Rightarrow x+y=167,5\)
Ta có:\(x^2+y^2=x^2+2xy+y^2-2xy=\left(x+y\right)^2-2xy=\left(167,5\right)^2-2.11=28034,25\)
Ta có: x2+y2
= (x+y)2-2xy
Thay x+y=-8 và xy=15, ta có:
x2+y2= 64-30
=> x2+y2=34