A)

Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )

\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)

\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)

\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)

Có:

\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)

\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)

B)

\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)

\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)

\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)

\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$

T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)

\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)

C)

\(2x=\sqrt{\frac{1-a}{a}}-\sqrt{\frac{a}{1-a}}\Rightarrow 4x^2=\frac{1-a}{a}+\frac{a}{1-a}-2\)

\(\Rightarrow 4(x^2+1)=\frac{1-a}{a}+\frac{a}{1-a}+2=(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}})^2\)

\(\Rightarrow \sqrt{4(x^2+1)}=\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}}\)

Khi đó:

\(C=\frac{2a\sqrt{4(1+x^2)}}{\sqrt{4(x^2+1)}-2x}=\frac{2a\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}} \right )}{\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}}-(\sqrt{\frac{1-a}{a}}-\sqrt{\frac{a}{1-a}})}=\frac{a\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}} \right )}{\sqrt{\frac{a}{1-a}}}\)

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

a: \(A=\left(\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)

\(=\dfrac{x+4\sqrt{x}+4-x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}}\)

\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}\)

c: 2x-3căn x-5=0

=>2x-5căn x+2căn x-5=0

=>2căn x-5=0

=>x=25/4

Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)

\(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

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

\(=x-\sqrt{x}+1\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^3+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\) \(\Rightarrow a+b=7\)

Bài 1: 

a: Ta có: \(x^2-2\sqrt{5}x+5=0\)

\(\Leftrightarrow x-\sqrt{5}=0\)

hay \(x=\sqrt{5}\)

b: Ta có: \(\sqrt{x+3}=1\)

\(\Leftrightarrow x+3=1\)

hay x=-2