Объяснение:
0\; ,\; y>0\; ,\; x\geq y\; \; \Rightarrow \; \; (x-y)\geq 0\; ,\; (x+y)\geq 0\; ;\\\\\sqrt{x^2-y^2}+\sqrt{x+y}=\sqrt{(x-y)(x+y)}+\sqrt{x+y}=\\\\=\sqrt{x-y}\cdot \sqrt{x+y}+\sqrt{x+y}=\sqrt{x+y}\cdot (\sqrt{x-y}+1)\\\\\\2)\; \; a>0\; ,\; b>0\; ,\; c>0\; \; \Rightarrow \; \; b+c>0\; ;\\\\\sqrt{ab+ac}-\sqrt{b^2+bc}=\sqrt{a\cdot (b+c)}-\sqrt{b\cdot (b+c)}=\\\\=\sqrt{a}\cdot \sqrt{b+c}-\sqrt{b}\cdot \sqrt{b+c}=\sqrt{b+c}\cdot (\sqrt{a}-\sqrt{b})" alt="1)\; \; x>0\; ,\; y>0\; ,\; x\geq y\; \; \Rightarrow \; \; (x-y)\geq 0\; ,\; (x+y)\geq 0\; ;\\\\\sqrt{x^2-y^2}+\sqrt{x+y}=\sqrt{(x-y)(x+y)}+\sqrt{x+y}=\\\\=\sqrt{x-y}\cdot \sqrt{x+y}+\sqrt{x+y}=\sqrt{x+y}\cdot (\sqrt{x-y}+1)\\\\\\2)\; \; a>0\; ,\; b>0\; ,\; c>0\; \; \Rightarrow \; \; b+c>0\; ;\\\\\sqrt{ab+ac}-\sqrt{b^2+bc}=\sqrt{a\cdot (b+c)}-\sqrt{b\cdot (b+c)}=\\\\=\sqrt{a}\cdot \sqrt{b+c}-\sqrt{b}\cdot \sqrt{b+c}=\sqrt{b+c}\cdot (\sqrt{a}-\sqrt{b})" align="absmiddle" class="latex-formula">