Question

помогите с 3,4,5. дам много балов​

Answer 1

Ответ:

Применяем формулы сокращённого умножения:

       $\bf (x-y)^2=(y-x)^2\ \ ,\ \ x^3+y^3=(x+y)(x^2-xy+y^2)\ \ ,$

       $\bf (x-y)(x+y)=x^2-y^2\ \ ,\ \ x\sqrt{x}=(\sqrt{x})^3$   .  

$\displaystyle 1)\ \ \Big(\frac{\sqrt{b}}{\sqrt{b}-\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{c}}\Big):\frac{\sqrt{b}}{\sqrt{b}-\sqrt{c}}=\frac{\sqrt{b}}{\sqrt{b}-\sqrt{c}}:\frac{\sqrt{b}}{\sqrt{b}-\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{c}}:\frac{\sqrt{b}}{\sqrt{b}-\sqrt{c}}=\\\\\\=1+\frac{\sqrt{b}}{\sqrt{c}}\cdot \frac{\sqrt{b}-\sqrt{c}}{\sqrt{b}}=1+\frac{\sqrt{b}-\sqrt{c}}{\sqrt{c}}=\frac{\sqrt{c}+\sqrt{b}-\sqrt{c}}{\sqrt{c}}=\frac{\sqrt{b}}{\sqrt{c}}$  

$\displaystyle 2)\ \ \Big(\frac{\sqrt{a}-\sqrt{b}}{a+\sqrt{ab}}-\frac{1}{a-b}\cdot \frac{(\sqrt{b}-\sqrt{a})^2}{\sqrt{a}+\sqrt{b}}\Big):\frac{\sqrt{a}-\sqrt{b}}{a+\sqrt{ab}}=\\\\\\=\frac{\sqrt{a}-\sqrt{b}}{a+\sqrt{ab}}:\frac{\sqrt{a}-\sqrt{b}}{a+\sqrt{ab}}-\frac{1}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}\cdot \frac{(\sqrt{a}-\sqrt{b})^2}{\sqrt{a}+\sqrt{b}}\cdot \frac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=$

$\displaystyle =1-\frac{\sqrt{a}\cdot (\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})^2}=1-\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}+\sqrt{b}-\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}$    

$\displaystyle 3)\ \ \frac{a\sqrt{a}+27}{\sqrt{a}-\sqrt{b}}\cdot \Big(\frac{\sqrt{a}-3}{a-3\sqrt{a}+9}-\frac{\sqrt{ab}-9}{a\sqrt{a}+27}\Big)=\\\\\\=\frac{(\sqrt{a}+3)(a-3\sqrt{a}+9)}{\sqrt{a}-\sqrt{b}}\cdot \frac{\sqrt{a}-3}{a-3\sqrt{a}+9}-\frac{a\sqrt{a}+27}{\sqrt{a}-\sqrt{b}}\cdot \frac{\sqrt{ab}-9}{a\sqrt{a}+27}=\\\\\\=\frac{(\sqrt{a}+3)(\sqrt{a}-3)}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{ab}-9}{\sqrt{a}-\sqrt{b}}=\frac{(a-9)-\sqrt{ab}+9}{\sqrt{a}-\sqrt{b}}=\frac{a-\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=$

$\displaystyle =\frac{\sqrt{a}\cdot (\sqrt{a}-\sqrt{b})}{\sqrt{a}-\sqrt{b}}=\sqrt{a}$