Question

Biết xy = 11 và \(x^2y+xy^2+x+y=10\) . Hãy tính \(x^2+y^2\)

Answer 1

ta có: x²y+xy²+x+y=(x²y+x)+(xy²+y)=x(xy+1)+y(xy+1)

=(x+y)(xy+1)=10

mà xy=11

=> x+y=\(\dfrac{5}{6}\)

Ta có: x²+y²=(x+y)²-2xy=\(\left(\dfrac{5}{6}\right)^2-2.11=-\dfrac{767}{36}\)

Answer 2

Bài giải
Ta có: $x^2+xy^2+x+y=10$
$<=>(xy+1)(x+y)=10$ mà $xy=11$, ta có:
$(xy+1)(x+y)=10$

$<=> 12.(x+y)=10$

$<=>x + y$ =\(\dfrac{10}{12}=\dfrac{5}{6}\)
Ta có:
$x^2+y^2=(x+y)^2 - 2xy$
=\(\left(\dfrac{5}{6}\right)^2-2.11\)
\(=\dfrac{25}{36}-2.11\\ =-\dfrac{767}{36}\)
Vậy \(x^2+y^2=-\dfrac{767}{36}\)