\(\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2a\right)\)

\(=a^2-c^2-\left(a-b\right)^2+c^2+b^2-2ab\)

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

\(=0\)

\(=\left(a^2-c^2\right)-\left(\left(a-b\right)^2-c^2\right)+b^2-2ab\)

\(=a^2-c^2-\left(a-b\right)^2+c^2+b^2-2ab\)

\(=\left(a^2-2ab+b^2\right)-\left(a-b\right)^2\)

\(=\left(a-b\right)^2-\left(a-b\right)^2=0\)

\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Đặt a+b=x ; b+c=y; c+a=z ta có:

\(x^3+y^3+z^3-3xyz\)

=\(\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)

=\(\left[\left(x+y\right)^3+z^3\right]-\left(3x^2y+3xy^2+3xyz\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

xong thay vào

1.

Cho $a+b+c=0$.

Ta có:

$a+b+c=0\Rightarrow a=-(b+c)$

$\Rightarrow a^2=(b+c)^2=b^2+c^2+2bc$

$\Rightarrow a^2-b^2-c^2=2bc$

Tương tự:

$b^2-c^2-a^2=2ca$

$c^2-a^2-b^2=2ab$

Do đó: $A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$

$=\dfrac{a^3+b^3+c^3}{2abc}$

Mà $a+b+c=0$ nên:

$a^3+b^3+c^3=3abc$

Suy ra $A=\dfrac{3abc}{2abc}$

$A=\dfrac32$

2.

Cho $abc=2$.

Ta có:

$A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}$

Vì $abc=2$ nên:

$ab=\dfrac2c,\quad bc=\dfrac2a,\quad ac=\dfrac2b$

Suy ra:

$A=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{2bc}{2+2bc+2b}$

$=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{bc}{1+bc+b}$

Quy đồng và sử dụng $abc=2$:

$A=1$

Đặt a+b = x ; b+c = y ; c+a = z

=> H = x^3 +y^3 +z^3 -3.x.y.z

         = [x+y]^3 -3.x^2.y -3.x.y^2+ z^3 - 3.x.y.z

          = {[x+y]^3+z^3} -3.x.y[x+y+z]

          = [x+y+z].{[x+y]^2-[x+y].z+z^2} +3.x.y[x+y+z]

          = [x+y+z] . [x^2+y^2+2.x.y-x.z-y.z+z^2+3.x.y]

           = [x+y+z]. [x^2+y^2+z^2-xy-y.z-x.z]

           = [a+b+b+c+c+a]. {[a+b]^2+[b+c]^2+[c+a]^2-[a+b].[b+c]-[a+b].[a+c] - [b+c].[c+a]}

            = 2.[a+b+c] .[a^2+b^2 +b^2 +c^2 +c^2 +a^2 +2.ab.+2.bc+2.ac-ab-b^2-ac-bc-a^2-ab-ac-bc-bc-c^2-ab-ac]

            = 2.[a+b+c].[a^2+b^2+c^2-ab-ac-bc]

rút gọn a

(-a-b+c)-(-a-b-c)