Question

Алгебра помогите 3 примера, я неспешу просто сделайте пж

Answer 1

Объяснение:

$\displaystyle \left(\frac{a-2b}{a^2+2ab}\stackrel{(2)}-\frac{1}{a^2-4b^2} \stackrel{(1)}:\frac{a+2b}{(2b-a)^2}\right)\stackrel{(3)}:\frac{a^2-2ab}{a^2+4ab+4b^2}$

$\displaystyle 1)\;\frac{1}{a^2-4b^2}:\frac{a+2b}{(2b-a)^2}=\frac{1}{(a-2b)(a+2b)}*\frac{(a-2b)^2}{a+2b} =\\\\=\frac{a-2b}{(a+2b)^2}$

$\displaystyle 2)\;\frac{a-2b}{a(a+2b)} ^{(a+2b}-\frac{a-2b}{(a+2b)^2}^{(a} =\\\\=\frac{a^2-4b^2-a^2+2ab}{a(a+2b)^2} =\frac{2b(a-2b)}{a(a+2b)^2}$

$\displaystyle 3)\;\frac{2b(a-2b)}{a(a+2b)^2}:\frac{a(a-2b)}{(a+2b)^2}=\frac{2b(a-2b)}{a(a+2b)^2} *\frac{(a+2b)^2}{a(a-2b)}=\frac{2b}{a^2}$

$\displaystyle \left(\frac{2a}{a+3}\stackrel{(1)}-\frac{4a}{a^2+6a+9}\right)\stackrel{(2)}*\frac{a^2-9}{a+1}\stackrel{(3)}-\frac{a^2-9a}{a+3}$

$\displaystyle 1)\;\frac{2a}{a+3} ^{(a+3}-\frac{4a}{(a+3)^2}^{(1}= \frac{2a^2+6a-4a}{(a+3)^2} =\\\\=\frac{2a^2+2a}{(a+3)^2}=\frac{2a(a+1)}{(a+3)^2}$

$\displaystyle 2)\; \frac{2a(a+1)}{(a+3)^2}*\frac{(a-3)(a+3)}{a+1} =\frac{2a(a-3)}{a+3}$

$\displaystyle 3)\;\frac{2a^2-6a}{a+3}-\frac{a^2-9a}{a+3} =\frac{2a^2-6a-a^2+9a}{a+3} =\\\\=\frac{a^2+3a}{a+3}=\frac{a(a+3)}{a+3}=a$

$\displaystyle \frac{\sqrt{a}-\sqrt{b} }{a+\sqrt{ab} } \stackrel{(3)}-\frac{1}{a-b}\stackrel{(1)}*\left(\frac{\sqrt{b}-\sqrt{a} }{\sqrt{a}+\sqrt{b} } \right) ^2\stackrel{(2)}:\frac{\sqrt{a}-\sqrt{b} }{a+\sqrt{ab} }$

$\displaystyle 1)\;\frac{1}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b} ) } *\frac{(\sqrt{a}-\sqrt{b} )^2 }{(\sqrt{a}+\sqrt{b})^2 } =\frac{\sqrt{a}-\sqrt{b} }{(\sqrt{a}+\sqrt{b})^3 }$

$\displaystyle 2)\;\frac{\sqrt{a}-\sqrt{b} }{(\sqrt{a}+\sqrt{b})^3 } :\frac{\sqrt{a}-\sqrt{b} }{\sqrt{a}(\sqrt{a}+\sqrt{b} ) }=\frac{\sqrt{a}-\sqrt{b} }{(\sqrt{a}+\sqrt{b})^3 } *\frac{\sqrt{a} (\sqrt{a}+ \sqrt{b}) }{\sqrt{a}-\sqrt{b} } =\\\\=\frac{\sqrt{a} }{(\sqrt{a}+\sqrt{b})^2 }$

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

Использовали формулы сокращенного умножения: