Question

1, Cho \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\) . Tính giá trị biểu thức \(Q=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)

2, Tìm nghiệm nguyên của phương trình :  \(2x^2+y^2+3xy+3x+2y+2=0\)

Answer 1

1)

xét a+b+c = (a+b+c)(\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)) = \(\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=\)

\(\frac{a^2}{b+c}+\frac{a\left(b+c\right)}{b+c}+\frac{b^2}{a+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c^2}{a+b}+\frac{c\left(a+b\right)}{a+b}=Q+a+b+c\)

<=> a+b+c =Q + a+b+c => Q=0

2) = (x+ y)² + (x+ 1)² +y(x+ 1) +x + y + 1 =0 <=> (x+ y)(x+ y+ 1) + (x+ 1)(x+ y+ 1) + 1= 0 <=> (x+ y+ 1)(2x+ y+ 1) = -1

=> \(\hept{\begin{cases}x+y+1=1\\2x+y+1=-1\end{cases}}\)hoặc \(\hept{\begin{cases}x+y+1=-1\\2x+y+1=1\end{cases}}\)<=> \(\hept{\begin{cases}x=-2\\y=2\end{cases}}\)hoặc \(\hept{\begin{cases}x=2\\y=-4\end{cases}}\)