Question

[Ngô Bảo An]

Gui8yuytutguytg

Answer 1

[Edogawa Conan]

1a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)

\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)

b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

2. a) ĐK: x \(\ge\)4

Ta có: \(\sqrt{16x-64}=2\) <=> \(4\sqrt{x-4}=2\) <=> \(\sqrt{x-4}=\frac{1}{2}\)

<=> \(x-4=\frac{1}{4}\) <=> \(x=\frac{17}{4}\left(tm\right)\)

b) \(16x^2-\left(1+\sqrt{2}\right)^2=0\) <=> \(\left(4x-1-\sqrt{2}\right)\left(4x+1+\sqrt{2}\right)=0\)

<=> \(\orbr{\begin{cases}x=\frac{1+\sqrt{2}}{4}\\x=\frac{-1-\sqrt{2}}{4}\end{cases}}\)

c)Đk: x \(\ge\)0

 \(x-2\sqrt{3x}+3=4\) <=> \(\left(\sqrt{x}-\sqrt{3}\right)^2=4\) 

<=> \(\left(\sqrt{x}-\sqrt{3}-2\right)\left(\sqrt{x}-\sqrt{3}+2\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+2\\\sqrt{x}=\sqrt{3}-2\left(loại\right)\end{cases}}\)

<=> \(x=5+4\sqrt{3}\)

Answer 2

[Edogawa Conan]

3a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{\sqrt{20}.\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}\)

\(=4\sqrt{5}-\frac{8\left(1+\sqrt{5}\right)}{5-1}=4\sqrt{5}-2\left(1+\sqrt{5}\right)=4\sqrt{5}-2-2\sqrt{5}=2\sqrt{5}-2\)

b) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}+9\sqrt{2}}\)

\(=\frac{2\left(2\sqrt{2}-3\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{6}\left(\sqrt{5}+3\sqrt{3}\right)}=-\frac{2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=-\frac{3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)