Giai hệ phương trình sau: $\sqrt[3]{x-y}$= $\sqrt{x-y}$ $x$+$y$=$\sqrt{x+y+2}$

Đáp án+Giải thích các bước giải:

$\begin{array}{l}\begin{cases}\sqrt[3]{x-y}=\sqrt[]{x-y}\\x+y=\sqrt[]{x+y+2}\\\end{cases}\\ĐKXĐ:\begin{cases}x \geq y\\x+y \geq -2\\\end{cases}\\HPT↔\begin{cases}(\sqrt[3]{x-y})^6=(\sqrt[]{x-y})^6\\(x+y)^2=x+y+2\\\end{cases}\\↔\begin{cases}(x-y)^2=(x-y)^3\\x^2+2xy+y^2=x+y+2\\\end{cases}\\↔\begin{cases}(x-y)^2(x-y-1)=0\\x^2+2xy+y^2=x+y+2\\\end{cases}\\↔\begin{cases}\left[ \begin{array}{l}x=y\\y=x-1\end{array} \right.\\x^2+2xy+y^2=x+y+2\\\end{cases}\\↔\left[ \begin{array}{l}\begin{cases}x=y\\x^2+2x^2+x^2=x+x+2\\\end{cases}\\\begin{cases}y=x-1\\x^2+2x(x-1)+(x-1)^2=x+x-1+2\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\begin{cases}x=y\\4x^2=2x+2\\\end{cases}\\\begin{cases}y=x-1\\x^2+2x^2-2x+x^2-2x+1=2x+1\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\begin{cases}x=y\\x^2-\dfrac{1}{2}x=\dfrac{1}{2}\\\end{cases}\\\begin{cases}y=x-1\\4x^2-6x=0\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\begin{cases}x=y\\x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}=\dfrac{9}{16}\\\end{cases}\\\begin{cases}y=x-y\\2x(2x-3)=0\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\begin{cases}x=y\\(x-\dfrac{1}{4})^2=\dfrac{9}{16}\\\end{cases}\\\begin{cases}y=x-1\\\left[ \begin{array}{l}x=0\\x=\dfrac{3}{2}\end{array} \right.\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\left[ \begin{array}{l}x=y=\dfrac{3+1}{4}=1(TM)\\x=y=\dfrac{-3+1}{4}=-\dfrac{1}{2}(TM)\end{array} \right.\\\left[ \begin{array}{l}\begin{cases}x=0(TM)\\y=x-1=-1(TM)\\\end{cases}\\\begin{cases}x=\dfrac{3}{2}\\y=x-1=\dfrac{1}{2}\\\end{cases}\end{array} \right.\end{array} \right.\\\text{vậy HPT có nghiệm}(x,y)=(1,1),(-\dfrac{1}{2},-\dfrac{1}{2}),(0,-1),(\dfrac{3}{2},\dfrac{1}{2})\\\end{array}$