Ta có: 

1) \(A=a\cdot b=\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\sqrt{9-5}=\sqrt{4}=2\)

2) \(B=a^2+b^2=\left(\sqrt{3+\sqrt{5}}\right)^2+\left(\sqrt{3-\sqrt{5}}\right)^2\)

\(=3+\sqrt{5}+3-\sqrt{5}=6\)

3) Xét: \(\left(a+b\right)^2=a^2+2ab+b^2=10\)

\(\Rightarrow a+b=\sqrt{10}\)

\(C=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=\sqrt{10}\cdot\left(6-2\right)\)

\(=4\sqrt{10}\)

4) \(a^5+b^5=\left(a+b\right)^5-\left(5a^4b+10a^3b^2+10a^2b^3+5ab^4\right)\)

\(=\left(\sqrt{10}\right)^5-5ab\left(a^3+b^3\right)-10a^2b^2\left(a+b\right)\)

\(=100\sqrt{10}-5\cdot2\cdot4\sqrt{10}-10\cdot2^2\cdot\sqrt{10}\)

\(=100\sqrt{10}-40\sqrt{10}-40\sqrt{10}\)

\(=20\sqrt{10}\)

1) đk: \(x\ge1\)

Ta có: \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)

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

\(\Leftrightarrow x-1=2x^2-2x\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)

Vậy x = 1

2) đk: \(x\ge\frac{1}{2}\)

Ta có: \(\sqrt{5x^2}=2x-1\)

\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)

\(\Leftrightarrow5x^2=4x^2-4x+1\)

\(\Leftrightarrow x^2+4x-1=0\)

\(\Leftrightarrow\left(x+2\right)^2-5=0\)

\(\Leftrightarrow\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{5}\left(ktm\right)\\x=-2-\sqrt{5}\left(ktm\right)\end{cases}}\)

=> PT vô nghiệm

3) đk: \(x\ge-1\)

Ta có: \(\sqrt{x+1}+\sqrt{9x+9}=4\)

\(\Leftrightarrow\sqrt{x+1}+3\sqrt{x+1}=4\)

\(\Leftrightarrow4\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=1\)

\(\Rightarrow x=0\)

4) đk: \(x\ge2\)

Ta có: \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)

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

\(\Leftrightarrow x-2=x\left(x-2\right)\)

\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

Vậy x = 2

6) đk: \(x\ge-\frac{7}{5}\)

Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)

\(\Leftrightarrow\frac{2x-3}{x-1}=2\)

\(\Leftrightarrow2x-3=2x-2\)

\(\Leftrightarrow0x=1\) vô lý

=> PT vô nghiệm

Đk: x \(\ge\)2013; y \(\ge\)2013

Ta có: A = \(\frac{\sqrt{x-2013}}{x}+\frac{\sqrt{y-2012}}{y+1}\ge0\forall x;y\)

(vì \(\sqrt{x-2013}\ge0\); x > 0; \(\sqrt{y-2012}\ge0\); y + 1 > 0)

Dấu "=" xảy ra <=> x - 2013 = 0 và y - 2012 = 0 <=> x = 2013 và y = 2012

Vậy MinA = 0 khi x =2013 và y = 2012

Ta lại có: A = \(\frac{\sqrt{x-2013}}{x}+\frac{\sqrt{y-2012}}{y+1}\le\frac{\frac{x-2013+1}{2}}{x}+\frac{\frac{y-2012+1}{2}}{y+1}\)(bđt cosi)

<=> A \(\le\frac{x-2012}{2x}+\frac{y-2011}{2\left(y+1\right)}=\frac{1}{2}-\frac{1006}{x}+\frac{y+1-2012}{2\left(y+1\right)}\)

<= > A \(\le\frac{1}{2}-\frac{1006}{x}+\frac{1}{2}-\frac{1006}{y+1}=1-1006\left(\frac{1}{x}+\frac{1}{y+1}\right)\)

Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (1)

CM bđt đúng: Từ (1) => (a + b)² > = 4ab <=> (a - b)² > = 0 (luôn đúng với mọi a,b > 0)

Khi đó: A \(\le1-\frac{1006.4}{x+y+1}=1-\frac{4024}{x+y+1}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{x-2013}=1\\\sqrt{y-2012}=1\\x=y+1\end{cases}}\) <=> \(\hept{\begin{cases}x=2014\\y=2013\end{cases}}\)

Vậy MaxA = \(1-\frac{4024}{2014+2013+1}=1-\frac{1006}{1007}=\frac{1}{1007}\) <=> x = 2014 và y = 2013

đk: \(x>0;x\ne9\)

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

b) Với x=0,25 ta có: \(P=\frac{\left(\sqrt{0,25}-1\right)^2}{\sqrt{0,25}}=0,5\)

c) \(P=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}-2=2-2=0\)

Dấu '=' xảy ra khi x=1 (tmdk). Vậy Min p =0 khi và chỉ khi x=1