Chăm học quá chị ạ

Dạ , bt làm ko giúp mk vs

Đợi tí mình làm cho

\(P=\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

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

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

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

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

\(=\left(x-z\right)\left[3y\left(y-z\right)-3x\left(y-z\right)\right]\)

\(=\left(x-z\right)\left(y-z\right)\left(3y-3x\right)\)

\(=3\left(x-z\right)\left(y-z\right)\left(y-x\right)\)

\(=3\left(z-x\right)\left(y-z\right)\left(x-y\right)\)

\(=3.2019\)

\(=6057\)

\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2\)

\(\Leftrightarrow M=2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2-2x^2y^2-x^2-y^2-1\)(Đặt M = .... cho gọn)

Có \(S=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)

\(\Rightarrow S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+M\)

\(\Rightarrow S^2=2x^2y^2+x^2+y^2+2019^2-2x^2y^2-x^2-y^2-1\)

\(\Rightarrow S=\sqrt{2019^2-1}\)

\(\sqrt{x^2-6x+9}-\sqrt{x^2-2x+1}=\sqrt{x^2}\)

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

\(\Rightarrow x-3-x+1-x=0\)

\(\Rightarrow-x=2\Rightarrow x=-2\)

Vậy......

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

\(\Leftrightarrow\left|x-3\right|-\left|x-1\right|-\left|x\right|=0\)

Xét \(x< 0\Leftrightarrow3-x+x-1+x=0\)

               \(\Leftrightarrow x=-2\)(tm)

Xét \(0\le x< 1\)\(\Leftrightarrow3-x+x-1-x=0\)

                         \(\Leftrightarrow x=1\left(l\right)\)

Xét \(1< x\le3\Leftrightarrow3-x-x+1-x=0\)

                      \(\Leftrightarrow4=3x\Leftrightarrow x=\frac{4}{3}\)(tm)

Xét \(x\ge3\Leftrightarrow x-3-x+1-x=0\)

              \(\Leftrightarrow x=-1\left(l\right)\)