a) \(P=\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-15\sqrt{x}+11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5x-7\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}\)

b) Ta có: \(P=\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{5\left(\sqrt{x}+3\right)-17}{\sqrt{x}+3}=5-\dfrac{17}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\)

\(\Rightarrow5-\dfrac{17}{\sqrt{x}+3}\ge-\dfrac{2}{3}\Rightarrow P_{min}=-\dfrac{2}{3}\) khi \(x=0\)

Vẽ đường cao AH \(\Rightarrow DE\parallel AH(\bot BC)\) mà D là trung điểm AB

\(\Rightarrow E\) là trung điểm BH \(\Rightarrow EB=EH\)

Ta có: \(EC^2-EB^2=\left(EC-EB\right)\left(EC+EB\right)=\left(EC-BH\right)BC\)

\(=CH.BC\)

tam giác ABC vuông tại A có đường cao AH nên áp dụng hệ thức lượng

\(\Rightarrow AC^2=CH.BC\Rightarrow AC^2=EC^2-EB^2\)

a) để căn thức có nghĩa thì \(3x^2+1\ge0\) (luôn đúng) nên căn luôn có nghĩa

b) để căn thức có nghĩa thì \(4x^2-4x+1\ge0\Rightarrow\left(2x-1\right)^2\ge0\) (luôn đúng)

nên căn luôn có nghĩa

c) để căn thức có nghĩa thì \(\dfrac{3}{x+4}\ge0\) mà \(3>0\Rightarrow x+4>0\Rightarrow x>-4\)

h) để căn thức có nghĩa thì \(x^2-4\ge0\Rightarrow x^2\ge4\Rightarrow\left|x\right|\ge2\)

i) để căn thức có nghĩa thì \(\dfrac{2+x}{5-x}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2+x\ge0\\5-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2+x\le0\\5-x< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-2\le x< 5\\\left\{{}\begin{matrix}x\le-2\\x>5\end{matrix}\right.\left(l\right)\end{matrix}\right.\Rightarrow-2\le x< 5\)

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

\(=\dfrac{3+\sqrt{5}}{9-5}-\dfrac{\sqrt{5}-1}{5-1}=\dfrac{3+\sqrt{5}}{4}-\dfrac{\sqrt{5}-1}{4}=\dfrac{4}{4}=1\)

\(\Delta'=\left(m+2\right)^2-\left(m^2+4m+3\right)=1>0\) 

\(\Rightarrow\) pt luôn có 2 nghiệm phân biệt

Áp dụng hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2m+4\\x_1x_2=m^2+4m+3\end{matrix}\right.\)

Ta có: \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=4\left(m+2\right)^2-2\left(m^2+4m+3\right)\)

\(=2m^2+8m+10=2\left(m^2+4m+4\right)+2=2\left(m+2\right)^2+2\ge2\)

\(\Rightarrow\) GTNN của \(x_1^2+x_2^2=2\) khi \(m=-2\)

\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{3\sqrt{x}-1}{1-x}\right)\left(\dfrac{2}{\sqrt{x}}-\dfrac{2}{x}\right)\left(x>0,x\ne1\right)\)

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

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

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

\(=\dfrac{6}{\sqrt{x}}\)

a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\)

b) \(23-8\sqrt{7}=4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2=\left(4-\sqrt{7}\right)^2\)

c) \(4-2\sqrt{3}=\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2=\left(\sqrt{3}-1\right)^2\)

d) \(11+6\sqrt{2}=3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(3+\sqrt{2}\right)^2\)

\(P=\left(1+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}+1}{x\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

\(\sqrt{20}=\sqrt{4.5}=\sqrt{2^2.5}=\left|2\right|.\sqrt{5}=2\sqrt{5}\)

a) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{\dfrac{6-2\sqrt{5}}{2}}+\sqrt{\dfrac{6+2\sqrt{5}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2.\sqrt{5}.1+1^2}{2}}+\sqrt{\dfrac{\left(\sqrt{5}\right)^2+2.\sqrt{5}.1+1^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{2}}=\dfrac{\sqrt{5}-1}{\sqrt{2}}+\dfrac{\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

b) \(2\sqrt{\dfrac{16}{5}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}=2\sqrt{\dfrac{4^2}{5}}-\sqrt{9.\dfrac{1}{27}}-6\sqrt{\left(\dfrac{2}{5}\right)^2.\dfrac{1}{3}}\)

\(=8\sqrt{\dfrac{1}{5}}-\sqrt{\dfrac{1}{3}}-\dfrac{12}{5}\sqrt{\dfrac{1}{3}}=\dfrac{8}{\sqrt{5}}-\dfrac{17}{5\sqrt{3}}=\dfrac{24\sqrt{5}-17\sqrt{3}}{15}\)

c) \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)

\(=\dfrac{\sqrt{8}-\sqrt{7}}{\left(\sqrt{8}+\sqrt{7}\right)\left(\sqrt{8}-\sqrt{7}\right)}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}=4\sqrt{7}\)

d) \(\sqrt{10-\sqrt{84}}-\sqrt{34+2\sqrt{189}}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.\sqrt{7}+\left(\sqrt{7}\right)^2}\)

\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{3}+\sqrt{7}\right)^2}=\sqrt{7}-\sqrt{3}-3\sqrt{3}-\sqrt{7}=-4\sqrt{3}\)