\(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

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

a, \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{17-2.3.2\sqrt{2}}-\sqrt{17+2.3.2\sqrt{2}}\)

\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}=\left|3-2\sqrt{2}\right|-\left|3+2\sqrt{2}\right|\)

\(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)

b, \(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)

\(=\sqrt{31-2.2.3\sqrt{3}}-\sqrt{31+2.2.3\sqrt{3}}\)

\(=\sqrt{\left(3\sqrt{3}-2\right)^2}-\sqrt{\left(3\sqrt{3}+2\right)^2}=\left|3\sqrt{3}-2\right|-\left|3\sqrt{3}+2\right|\)

\(=3\sqrt{3}-2-3\sqrt{3}-2=-4\)

để \(\frac{1}{\sqrt{4-x}}\text{ có nghĩa thì }\hept{\begin{cases}4-x\ge0\\4-x\ne0\end{cases}\Leftrightarrow x< 4}\)

\(\frac{4x}{\sqrt{x-1}}\text{ có nghĩa thì }\hept{\begin{cases}x-1\ge0\\x-1\ne0\end{cases}\Leftrightarrow x>1}\)

Ta có \(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}=\frac{abc}{a^2}+\frac{abc}{b^2}+\frac{abc}{c^2}=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

Khi đó \(abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge a+b+c\)

<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{a+b+c}{abc}\)

<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}\)

<=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\ge\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}\)

<=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ab}\ge0\)

<=> \(\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)\ge0\)

<=> \(\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2\ge0\left(\text{đúng }\forall a;b;c>0\right)\)

=> ĐPCM (Dấu "=" xảy ra <=> a = b = c) 

ta có :

\(\hept{\begin{cases}AB^2=BD.BC=9\left(9+16\right)=225\\AC^2=CD.CB=16\left(16+9\right)=400\end{cases}}\Leftrightarrow\hept{\begin{cases}AB=15\\AC=20\end{cases}}\)

nên diện tích ABC là : \(\frac{1}{2}AB.AC=\frac{1}{2}.15.20=150cm^2\)

Với x > = 0 ; \(x\ne1\)

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

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

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

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