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`)

B1 : a, M = x³-3xy(x-y)-y³-x²+2xy-y²

= ( x³-y³)-3xy(x-y) -(x²-2xy+y²)

= (x-y)(x²+xy+y²)-3xy(x-y)-(x-y)²

= (x-y) [(x²+xy+y²-3xy-(x-y)]

= (x-y)[(x²-2xy+y²)-(x-y)

= (x-y)[(x-y)²-(x-y)]

= (x-y)(x-y)(x-y-1)

= (x-y)²(x-y-1)

= 7²(7-1) = 49 . 6= 294

N = x²(x+1)-y²(y-1)+xy-3xy(x-y+1)-95

= x³+x²-(y³-y²)+xy-(3x²y-3xy²+3xy)-95

= x³+x²-y³+y²+xy-3x²y+3xy²-3xy-95

= (x³-y³)+(x²-2xy+y²)-(3x²y+y²)-(3x²y-3xy²)-95

=(x-y)(x²+xy+y²)+(x-y)²-3xy(x-y)-95

= (x-y)(x²+xy+y²+x-y-3xy)-95

= (x-y)[(x²-2xy+y²)+(x-y)]-95

= (x-y)[(x-y)²+(x-y)]-95

=(x-y)(x-y)(x-y+1)-95

= (x-y)²(x-y+1)-95

= 7²(7+1)-95=297

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x+y+z}{y+z+x}=\dfrac{x+y+z}{x+y+z}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
Do đó \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)
Thay vào biểu thức \(P=\left(x-y\right)^{2022}+\left(y-z\right)^{2023}+\left(x-z-1\right)^{202}\),ta có:
\(P=0^{2022}+0^{2023}+\left(-1\right)^{202}\)
\(=0+0+1\)
\(=1\)

a)<=>

A,=(x+y)(x-y)=x^2-y^2

x=(-1/2)^5:(1/2)^4=-1/2

x^2=1/4

y=8^2/(-2)^5=-2

y^2=4

A=1/4-4=-15/4