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Biết xy=11 và x2y+xy2+x+y=2010.Tính x2+y2
ta có:x2y+xy2+x+y=2010
<=>xy(x+y)+x+y=2010
<=>(x+y)(xy+1)=2010
<=>x+y=167,5
<=>(x+y)2=x2+y2+2xy=28056,25
<=>x2+y2=28056,25-22=28034,25
\(x^2y+xy^2+x+y=2010\)
\(\Leftrightarrow xy\left(x+y\right)+x+y=2010\)
\(\Leftrightarrow\left(x+y\right)\left(xy+1\right)=2010\)
\(\Leftrightarrow\left(x+y\right)\left(11+1\right)=2010\)
\(\Leftrightarrow x+y=\frac{2010}{11+1}=\frac{332}{5}\)
Ta có \(x^2+y^2=\left(x+y\right)^2-2xy=\left(\frac{332}{5}\right)^2-2.11=\frac{112137}{4}\)
a ) x ^ 2 + 2xy + 7x + 7y + y ^2 + 10 = ( x + y ) ^2 + 7 ( x + y ) + 10 = ( x + y ) ( x + y + 17 )
Theo bài ra ta có:
\(x^2y+xy^2+x+y=2010\)
\(\Rightarrow xy\left(x+y\right)+\left(x+y\right)=2010\)
\(\Rightarrow\left(x+y\right)\left(xy+1\right)=2010\)
\(\Rightarrow\left(x+y\right)\left(11+1\right)=2010\)
\(\Rightarrow12\left(x+y\right)=2010\Rightarrow x+y=2010\div12=167,5\)
Ta có: \(A=x^4+y^4=\left(x^2\right)^2+2x^2y^2+\left(y^2\right)^2-2x^2y^2\)
\(=\left(x^2+y^2\right)^2-2\left(xy\right)^2\)
\(=\left[\left(x+y\right)^2-2xy\right]^2-2\times11^2\)
\(\Rightarrow\left[\left(167,5\right)^2-2.11\right]^2-245\)
\(\Rightarrow\left(28056,25-22\right)^2-245=785918928,0625\)
\(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\Rightarrow x=y=z\)
Ta lại có : \(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)
\(\Rightarrow3x^{2009}=3^{2010}\Rightarrow x^{2009}=3^{2009}\Rightarrow x=3\)
\(\Rightarrow x=y=z=3\)
Vậy .............
a) \(A=x^2y+y+xy^2-x\) (hẳn đề là vậy)
\(A=xy\left(x+y\right)+\left(y-x\right)\)
\(A=\left(-5\right).2\left(-5+2\right)+2+5\)
\(A=30+7=37\)
b) \(B=3x^3-2y^3-6x^2y^2+xy\)
\(B=3.\left(\frac{2}{3}\right)^3-2.\left(\frac{1}{2}\right)^3-6.\left(\frac{2}{3}\right)^2.\left(\frac{1}{2}\right)^2+\frac{2}{3}.\frac{1}{2}\)
\(B=\frac{8}{9}-\frac{1}{4}-\frac{2}{3}+\frac{1}{3}\)
\(B=\frac{11}{36}\)
c) \(C=2x+xy^2-x^2y-2y\)
\(C=2.\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right).\left(-\frac{1}{3}\right)^2-\left(-\frac{1}{2}\right)^2.\left(-\frac{1}{3}\right)-2.\left(-\frac{1}{3}\right)\)
\(C=-1-\frac{1}{18}+\frac{1}{12}+\frac{2}{3}\)
\(C=-\frac{11}{36}\)
Sửa đề
\(2A=2x^2+2y^2+2xy-2x+2y+2\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)\)
\(=\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
\(\Rightarrow A_{min}=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(x^2y+xy^2+x+y=xy\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(xy+1\right)=12\left(x+y\right)=2010\)
\(\Rightarrow x+y=\dfrac{2010}{12}\)
\(\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy=\left(\dfrac{2010}{12}\right)^2-2\cdot11=\dfrac{112137}{4}\)