Question

[Hứa Thập Ca]

Giúp mình với ạ

Answer 1

[Yen Nhi]

Answer:

Bài 1:

a. Ta xét vế trái:

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

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

\(=\sqrt{5}+3+\sqrt{3}-\sqrt{5}-3\)

\(=\sqrt{3}\)

b. Với \(a\ge1\)

\(P=a-\left(\frac{1}{\sqrt{a}-\sqrt{a-1}}-\frac{1}{\sqrt{a}+a-1}\right)\)

\(=a-\frac{\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}\)

\(=a-\frac{2\sqrt{a-1}}{a-a+1}\)

\(=a-\frac{2\sqrt{a-1}}{1}\)

\(=a-2\sqrt{a-1}\)

\(=a-1-2\sqrt{a-1}+1\)

\(=\left(\sqrt{a-1}-1\right)\ge0\forall a\ge1\)

\(\Rightarrow P\ge0\)

Answer 2

[Yen Nhi]

Answer:

Bài 2:

\(3x+\sqrt{2}=2\left(x+\sqrt{2}\right)\)

\(\Rightarrow3x+\sqrt{2}=2x+2\sqrt{2}\)

\(\Rightarrow3x-2x=2\sqrt{2}-\sqrt{2}\)

\(\Rightarrow x=\sqrt{2}\)

\(3\sqrt{x-2}-\sqrt{x^2-4}=0\left(ĐK:-2\le x\le2\right)\)

\(\Rightarrow3\sqrt{x-2}=\sqrt{x^2-4}\)

\(\Rightarrow9\left(x-2\right)=x^2-4\)

\(\Rightarrow9x-2-x^2+4=0\)

\(\Rightarrow-x^2+9x+2=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9+\sqrt{89}}{2}\text{(Loại)}\\x=\frac{9-\sqrt{89}}{2}\end{cases}}\)