ta có : \(x^2y+xy^2+x+y=2010\)(1)

\(\Leftrightarrow xy\times\left(x+y\right)+\left(x+y\right)=2010\)

\(\Leftrightarrow\)( xy + 1 ) ( x + y ) = 2010

mà xy=11 \(\Rightarrow\)xy+1=12

(1)\(\Leftrightarrow\)12 (x + y ) = 2010

     \(\Leftrightarrow\)x + y = 167,5

lại có S\(=x^3+y^3\)

         S \(=\left(x+y\right)^3-3xy\left(x+y\right)\)

         S\(=167,5^3-3\times11\times167,5\)

         S \(=\)4693894,375

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 \(x^2+y^2=\frac{112137}{4}.\)

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

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:\(x^2+y^2=x^2+2xy+y^2-2xy=\left(x+y\right)^2-2xy=\left(167,5\right)^2-2.11=28034,25\)

k mk đi làm ơn 

mk đang bị âm điểm

bạn giúp mình đi làm ơn

mình đang ko biết cách làm

a.)2M=2x^2+2y^2-2xy-2x-2y+2

2M=(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)

2M=(x-y)^2+(x-1)^2+(y-1)^2>=0

Min 2M=0

Min M =0

Dấu = xảy ra khi x=y=1 

b.)x^2y+xy^2+x+y=2010

xy(x+y)+x+y=2010

(x+y)(xy+1)=2010

(x+y)2=2010

x+y=1005

(x+y)^2=1005^2

x^2+2xy+y^2=1005^2

x^2+y^2=1005^2-2xy

x^2+y^2=1005^2-2
 

\(x^2+2xy+7x+7y+y^2+10\)

\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+\frac{49}{4}-\frac{9}{4}\)

\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)

\(=\left(x+y+2\right)\left(x+y+5\right)\)

b)Ta có: x²y+xy²+x+y=2010

<=>xy.x+xy.y+x+y=2010

<=>11x+11y+x+y=2010

<=>12(x+y)=2010

<=>x+y=167,5

=>(x+y)²=28056,25

<=>x²+y²+2xy=28056,25

<=>x²+y²=28034,25

\(\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)

\(=\left(x+y\right)^2+2\left(x+y\right).\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)   (áp dụng HĐT số 1)

\(=\left(x+y+\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2\)  

\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)(áp dụng HĐT số 3)

\(=\left(x+y+2\right)\left(x+y+5\right)\)

\(x^2y+xy^2+x+y=2010\)

\(\Leftrightarrow xy\left(x+y\right)+x+y=2010\)

\(\Leftrightarrow\left(x+y\right)\left(xy+1\right)=2010\)

\(\Leftrightarrow\left(x+y\right)\left(11+1\right)=2010\)

\(\Leftrightarrow x+y=\frac{2010}{11+1}=\frac{332}{5}\)

Ta có  \(x^2+y^2=\left(x+y\right)^2-2xy=\left(\frac{332}{5}\right)^2-2.11=\frac{112137}{4}\)