Lời giải:
BĐT cần cm tương đương với:
$3(x+y+1)^2+1-3xy\geq 0$

$\Leftrightarrow 3x^2+3y^2+3xy+6x+6y+4\geq 0$

$\Leftrightarrow 12x^2+12y^2+12xy+24x+24y+16\geq 0$

$\Leftrightarrow 3(4x^2+y^2+4xy)+9y^2+24x+24y+16\geq 0$

$\Leftrightarrow 3(2x+y)^2+12(2x+y)+9y^2+12y+16\geq 0$

$\Leftrightarrow 3[(2x+y)^2+4(2x+y)+4]+(9y^2+12y+4)\geq 0$

$\Leftrightarrow 3(2x+y+2)^2+(3y+2)^2\geq 0$ (luôn đúng)

Do đó ta có đpcm.

Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2}\right)\left(x-\sqrt{x^2+2}\right)\left(y-1+\sqrt{y^2-2y+3}\right)=2\left(x-\sqrt{x^2+2}\right)\)

\(\Leftrightarrow-2\left(y-1+\sqrt{y^2-2y+3}\right)=2\left(x-\sqrt{x^2+2}\right)\)

\(\Leftrightarrow x-\sqrt{x^2+2}+y-1+\sqrt{y^2-2y+3}=0\) (*)

\(\left(x+\sqrt{x^2+2}\right)\left(y-1+\sqrt{y^2-2y+3}\right)=2\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2}\right)\left(y-1+\sqrt{y^2-2y+3}\right)\left(y-1-\sqrt{y^2-2y+3}\right)=2\left(y-1-\sqrt{y^2-2y+3}\right)\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2}\right).-2=2\left(y-1-\sqrt{y^2+2y+3}\right)\)

\(\Leftrightarrow y-1-\sqrt{y^2+2y+3}+x+\sqrt{x^2+2}=0\) (2*)

Cộng vế với vế của (*) và (2*) => \(2x+2y-2=0\)

\(\Leftrightarrow x+y=1\)

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\)

\(\Leftrightarrow x^3+y^3+3xy=1\)

Ta có:`(x+sqrt{x^2+2})(sqrt{x^2+2}-x)=2`

`<=>sqrt{x^2+2}-x=y-1+sqrt{y^2-2y+3}`

`<=>sqrt{x^2+2}-sqrt{y^2-2y+3}=x+y-1(1)`

CMTT:`sqrt{y^2-2y+3}-(y-1)=x+sqrt{x^2+2}`

`<=>sqrt{y^2-2y+3}-y+1=x+sqrt{x^2+2}`

`<=>sqrt{y^2-2y+3}-sqrt{x^2+2}=x+y-1(2)`

Cộng từng vế (1)(2) ta có:

`2(x+y-1)=0`

`<=>x+y-1=0`

`<=>x+y=1`

`<=>(x+y)^3=1`

`<=>x^3+y^3+3xy(x+y)=1`

`<=>x^3+y^3+3xy=1`(do `x+y=1`)

Áp dụng HĐT :(a-b)³ =a ³-3a²b+3ab² -b³

                       => a³ -b³ = (a-b)³ +3ab(a-b)

Biến đổi vế phải: x³ -y³ = (x-y) ³ + 3xy(x-y)

               = 1+3xy = Vế trái     (vì x-y=1)(đpcm)

Ta có:

x³-y³=(x-y)(x²+xy+y²)

       =1(x²-2xy+y²+3xy)

       =(x-y)²+3xy

       =1+3xy => ĐPCM

x^3+3xy+y^3

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

=1-3xy+3xy

=1

Đặt B=x³+y³=1-3xy

Ta có (x+y)³=x³+y³+3x²y+3xy² 

<=>(x+y)³=x³+y³+3xy(x+y)

Mà x+y=1 nên

1=x³+y³+3xy.1

Vậy B=1 

chứng minh ko phải tính bạn ơi

Bến đổi VT ta được :

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

\(=\left(x+y\right)^2-3xy=1-3xy=VP\)(đpcm)

\(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}=\dfrac{1}{x-y}\)

\(VT=\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)

\(=\dfrac{2x^2+2xy+xy+y^2}{\left(2x^3+x^2y\right)+\left(-2xy^2-y^3\right)}\)

\(=\dfrac{\left(2x^2+2xy\right)+\left(xy+y^2\right)}{x^2\left(2x+y\right)-y^2\left(2x+y\right)}\)

\(=\dfrac{2x\left(x+y\right)+y\left(x+y\right)}{\left(x^2-y^2\right)\left(2x+y\right)}\)

\(=\dfrac{\left(2x+y\right)\left(x+y\right)}{\left(x^2-y^2\right)\left(2x+y\right)}\)

\(=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{1}{x-y}=VP\left(đpcm\right)\)