Question

Дуже потрібно!!!
6 и 7 завдання

Answer 1

$\displaystyle\bf\\6)\\\\\Big(\frac{1}{x^{\frac{1}{2} } } -x^{\frac{1}{2} } \Big)\Big(\frac{x^{\frac{1}{2} } -1}{x^{\frac{1}{2} }+1 } -\frac{x^{\frac{1}{2} } +1}{x^{\frac{1}{2} }-1 } \Big)=\\\\\\=\frac{1-x^{\frac{1}{2} } \cdot x^{\frac{1}{2} } }{x^{\frac{1}{2} } } \cdot\frac{(x^{\frac{1}{2} }-1)\cdot(x^{\frac{1}{2} } -1)-(x^{\frac{1}{2} } +1)\cdot(x^{\frac{1}{2} } +1)}{(x^{\frac{1}{2} } +1)\cdot(x^{\frac{1}{2} } -1)} =$

$\displaystyle\bf\\=\frac{1-(x^{\frac{1}{2} } )^{2} }{x^{\frac{1}{2} } } \cdot\frac{(x^{\frac{1}{2} } )^{2} -2x^{\frac{1}{2} } +1-(x^{\frac{1}{2} } )^{2} -2x^{\frac{1}{2} } -1}{(x^{\frac{1}{2} } )^{2} -1} =\\\\\\=\frac{1-x}{x^{\frac{1}{2} } } \cdot \frac{-4x^{\frac{1}{2} } }{x-1} =\frac{1-x}{x^{\frac{1}{2} } } \cdot \frac{4x^{\frac{1}{2} } }{1-x} =4\\\\\\Otvet \ : \ 4$

$\displaystyle\bf\\7)\\\\ODZ \ : \\\\\left \{ {{x+21\geq 0} \atop {x-3\geq 0}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x\geq -21} \atop {x\geq 3} \right. \ \ \ \Rightarrow \ \ \ \boxed{x\geq 3}\\\\\\\sqrt{x+21} +\sqrt{x-3} =6\\\\\\\sqrt{x+21} =6-\sqrt{x-3} \\\\\\(\sqrt{x+21})^{2} =(6-\sqrt{x-3} )^{2} \\\\\\x+21=36-12\sqrt{x-3} +x-3\\\\\\x+12\sqrt{x-3} -x=33-21\\\\\\12\sqrt{x-3} =12\\\\\\\sqrt{x-3} =1\\\\\\x-3=1\\\\\\x=4\\\\\\Otvet \ : \ 4$