Giúp mih vs ậbnansbsjanabvscsv

Giải thích các bước giải:

Bài 8:

a.Ta có:

$\begin{aligned}
A &= \left( \dfrac{1}{\sqrt{a}-1} - \dfrac{2\sqrt{a}}{a\sqrt{a} + \sqrt{a} - a - 1} \right) : \left( 1 - \dfrac{2\sqrt{a}}{a+1} \right) \\
&= \left( \dfrac{1}{\sqrt{a}-1} - \dfrac{2\sqrt{a}}{\sqrt{a}(a+1) - (a+1)} \right) : \left( \dfrac{a+1 - 2\sqrt{a}}{a+1} \right) \\
&= \left( \dfrac{1}{\sqrt{a}-1} - \dfrac{2\sqrt{a}}{(a+1)(\sqrt{a}-1)} \right) : \left( \dfrac{(\sqrt{a}-1)^2}{a+1} \right) \\
&= \left( \dfrac{a+1}{(a+1)(\sqrt{a}-1)} - \dfrac{2\sqrt{a}}{(a+1)(\sqrt{a}-1)} \right) \cdot \dfrac{a+1}{(\sqrt{a}-1)^2} \\
&= \dfrac{a - 2\sqrt{a} + 1}{(a+1)(\sqrt{a}-1)} \cdot \dfrac{a+1}{(\sqrt{a}-1)^2} \\
&= \dfrac{(\sqrt{a}-1)^2}{(a+1)(\sqrt{a}-1)} \cdot \dfrac{a+1}{(\sqrt{a}-1)^2} \\
&= \dfrac{\sqrt{a}-1}{a+1} \cdot \dfrac{a+1}{(\sqrt{a}-1)^2} \\
&= \dfrac{1}{\sqrt{a}-1}
\end{aligned}$

b.Ta có:

$\begin{aligned}
\dfrac{a}{4+2\sqrt{3}} &= \dfrac{2\sqrt{2-\sqrt{3}}}{\sqrt{6}-\sqrt{2}} \\
\dfrac{a}{4+2\sqrt{3}} &= \dfrac{\sqrt{4(2-\sqrt{3})}}{\sqrt{2}(\sqrt{3}-1)} \\
\dfrac{a}{4+2\sqrt{3}} &= \dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}(\sqrt{3}-1)} \\
\dfrac{a}{4+2\sqrt{3}} &= \dfrac{\sqrt{(\sqrt{6}-\sqrt{2})^2}}{\sqrt{6}-\sqrt{2}} \\
\dfrac{a}{4+2\sqrt{3}} &= \dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} \\
\dfrac{a}{4+2\sqrt{3}} &= 1 \\
a &= 4 + 2\sqrt{3} \\
a &= (\sqrt{3} + 1)^2 \quad (\text{Thỏa mãn ĐK } 0 \le a \ne 1)
\end{aligned}$

Suy ra:

$\begin{aligned}
A &= \dfrac{1}{\sqrt{a}-1} \\
A &= \dfrac{1}{\sqrt{(\sqrt{3}+1)^2}-1} \\
A &= \dfrac{1}{(\sqrt{3}+1)-1} \\
A &= \dfrac{1}{\sqrt{3}} \\
A &= \dfrac{\sqrt{3}}{3}
\end{aligned}$

Bài 9:

a.Ta có:

$\begin{aligned}
A &= \left( 1 - \dfrac{2\sqrt{a}}{a+1} \right) : \left( \dfrac{1}{\sqrt{a}+1} - \dfrac{2\sqrt{a}}{a\sqrt{a} + \sqrt{a} + a + 1} \right) \\
&= \dfrac{a+1 - 2\sqrt{a}}{a+1} : \left( \dfrac{1}{\sqrt{a}+1} - \dfrac{2\sqrt{a}}{\sqrt{a}(a+1) + (a+1)} \right) \\
&= \dfrac{(\sqrt{a}-1)^2}{a+1} : \left( \dfrac{1}{\sqrt{a}+1} - \dfrac{2\sqrt{a}}{(a+1)(\sqrt{a}+1)} \right) \\
&= \dfrac{(\sqrt{a}-1)^2}{a+1} : \dfrac{a+1 - 2\sqrt{a}}{(a+1)(\sqrt{a}+1)} \\
&= \dfrac{(\sqrt{a}-1)^2}{a+1} : \dfrac{(\sqrt{a}-1)^2}{(a+1)(\sqrt{a}+1)} \\
&= \dfrac{(\sqrt{a}-1)^2}{a+1} \cdot \dfrac{(a+1)(\sqrt{a}+1)}{(\sqrt{a}-1)^2} \\
&= \sqrt{a} + 1
\end{aligned}$

b.Ta có:

$\begin{aligned}
a &= 2026 - 2\sqrt{2025} \\
&= 2025 - 2\sqrt{2025} + 1 \\
&= (\sqrt{2025} - 1)^2 \\
&= (45 - 1)^2 \\
&= 44^2 \\
\Rightarrow A &= \sqrt{44^2} + 1 \\
&= 44 + 1 \\
&= 45
\end{aligned}$

Bài 10: 

Ta có:

$\begin{aligned}
a &= \sqrt{\dfrac{5}{3}} + \sqrt{\dfrac{3}{5}} \\
&= \dfrac{\sqrt{5}}{\sqrt{3}} + \dfrac{\sqrt{3}}{\sqrt{5}} \\
&= \dfrac{5 + 3}{\sqrt{15}} \\
&= \dfrac{8}{\sqrt{15}}
\end{aligned}$

Suy ra:

$\begin{aligned}
A &= \sqrt{15a^2 - 8a\sqrt{15} + 16} \\
&= \sqrt{(\sqrt{15}a)^2 - 2 \cdot \sqrt{15}a \cdot 4 + 4^2} \\
&= \sqrt{(\sqrt{15}a - 4)^2} \\
&= \left| \sqrt{15}a - 4 \right| \\
&= \left| \sqrt{15} \cdot \dfrac{8}{\sqrt{15}} - 4 \right| \\
&= |8 - 4| \\
&= 4
\end{aligned}$