a ) x ^ 2 + 2xy + 7x + 7y + y ^2 + 10 = ( x + y ) ^2 + 7  ( x + y ) + 10 = ( x + y ) ( x + y + 17 )

bạn ơi còn phần b

b, x²y+xy²+x+y=2010

=>11x+11y+(x+y)=2010

=>11(x+y)+(x+y)=2010

12(x+y)=2010

=>x+y=167,5

=>(x+y)²=28056,25

=>x²+2xy+y²=28056,25

=>x²+y²=28034,25

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 \(x^2+y^2=\frac{112137}{4}.\)

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

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

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

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

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

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

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)

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

Kết quả cuối cùng bằng:(x+y).(x+y+7)
 

\(\left(x+y\right)^2+7\left(x+y\right)+10=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10=\left(x+y+2\right)\left(x+y+5\right)\)

tui   bit làm rùi khỏi đey]

(y+x+2)*(y+x+5)

chả biết có đúng ko

\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10=\left(x+y\right)^2+7\left(x+y\right)+10\)

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

^^