Question

[Phạm Mai Phương]

Giải hệ phương trình

{x.(x+1)+y.(y-1)=6 và x+y=3

Answer 1

[Tuii cũg dễx thưng màk]

Ta có :

- Xét x+y=3 \(\Rightarrow\hept{\begin{cases}x=3-y\\\left(x+y\right)^2=9\end{cases}}\)

\(x\left(x+1\right)+y\left(y-1\right)=6\)

\(\Rightarrow x^2+x+y^2-y=6\)

\(\Rightarrow\left(x^2+2xy+y^2\right)-2xy+x-y=6\)

\(\Rightarrow\left(x+y\right)^2-6=2xy-x+y\)

\(\Rightarrow9-6=2xy-x+y\)

\(\Rightarrow2xy-x+y=3\)

\(\Rightarrow2\left(3-y\right)y-\left(3-y\right)+y=3\)

\(\Rightarrow6y-2y^2-3+y+y=3\)

\(\Rightarrow6y+2y-2y^2=3+3\)

\(\Rightarrow8y-2y^2=6\)

\(\Rightarrow2y\left(4-y\right)=6\)

\(\Rightarrow4y-y^2=3\)

\(\Rightarrow4y-y^2=x+y\)

\(\Rightarrow x=3y-y^2=y\left(3-y\right)\)

\(\Rightarrow x=y\left(3-3+x\right)\)

\(\Rightarrow x=xy\)

\(\Rightarrow y=1\)

\(\Rightarrow x=3-y=3-1=2\)

Vậy x=2 ; y=1.

 Nhớ kb với mình nha!