Ta có

$x+1=\frac{b^2+c^2-a^2}{2bc}+1=\frac{b^2+2bc+c^2-a^2}{2bc}=\frac{(b+c{)}^2-a^2}{2bc}$

Suy ra

$y(x+1)=\frac{a^2-(b-c{)}^2}{(b+c{)}^2-a^2}.\frac{(b+c{)}^2-a^2}{2bc}=\frac{a^2-(b-c{)}^2}{2bc}$

Do đó

$P=x+y+xy=x+y(x+1)=\frac{b^2+c^2-a^2}{2bc}+\frac{a^2-(b-c{)}^2}{2bc}=\frac{b^2+c^2-a^2+a^2-(b-c{)}^2}{2bc}=1$

Lời giải:

Xét hiệu:

$a^2+b^2+c^2-(2ab-2ac+2bc)=a^2+b^2+c^2-2ab+2ac-2bc$

$=(a^2+b^2-2ab)+c^2+2c(a-b)$

$=(a-b)^2+c^2+2c(a-b)=(a-b+c)^2\geq 0, \forall a,b,c\in\mathbb{R}$
$\Rightarrow a^2+b^2+c^2\geq 2ab-2ac+2bc$

Vậy ta có đpcm.

Dấu "=" xảy ra khi $a-b+c=0$

Bạn lưu ý lần sau gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để đề được rõ ràng hơn nhé.

(x+1)(y+1)=xy+x+y+1 => P=xy+x+y= ( x+1)(y+1)-1

\(\left(x+1\right)=\dfrac{\left(b+c\right)^2-a^2}{2bc}=\dfrac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}\)

\(\left(y+1\right)=\dfrac{a^2-\left(b-c\right)^2+\left(b+c\right)^2-a^2}{\left(b+c+a\right)\left(b+c-a\right)}=\dfrac{4bc}{\left(b+c+a\right)\left(b+c-a\right)}\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=\dfrac{4bc}{2bc}=2=>xy+x+y=2-1=1\)

2) b)

Do \(a+b+c=9\Rightarrow\left(a+b+c\right)^2=81\) 

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=81\)

\(\Rightarrow2\left(ab+bc+ac\right)=81-141=-60\)

\(ab+bc+ac=-60:2=-30\)

a, B=x^3 + 3xy +y^3 = x^3 +3xy(x+y)+y^3 (vì x+y=1)

                           = (x+y)^3

                           = 1^3 =1

b, (a+b+c)^2 =a^2 +b^2 +c^2 +2ab +2bc +2ac

    9^2 = 141 +2(ab+bc+ac)

    -60 = 2(ab+bc+ac)

    ab+ac+bc=-30

Vậy M=-30

c, N =(x+y)^3 -3(x+y)(x^2+y^2) +2(x^3+y^3)

       = x^3 + 3x^2 .y + 3xy^2 + -3(x^3+xy^2 +x^2 .y+y^3)+ 2x^3 +2y^3

       = x^3 +3x^2 .y + 3xy^2 - 3x^3 -3xy^2 -3x^2 .y -3y^3 +2x^3 +2y^3

       = 0

Vậy N=0 .Chúc bạn học tốt.