bổ sung thêm là loại trường hợp \(x=\dfrac{1-\sqrt{3}}{2}\) nha bạn do \(x\ge0\)

nên chỉ có nghiệm là \(x=\dfrac{1+\sqrt{3}}{2}\)

\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\Rightarrow2x-\sqrt{x}=\sqrt{x}+1\Rightarrow2x-2\sqrt{x}-1=0\)

\(\Delta=\left(-2\right)^2-4.\left(-1\right).2=12\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{2-\sqrt{12}}{4}=\dfrac{1-\sqrt{3}}{2}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{2+\sqrt{12}}{4}=\dfrac{1+\sqrt{3}}{2}\end{matrix}\right.\)

hình bên trái: 

Ta có: \(BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+\left(7,5\right)^2}=\dfrac{17}{2}\left(cm\right)\)

Ta có: \(\left\{{}\begin{matrix}AB^2=BH.BC\\AC^2=CH.BC\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4^2=x.\dfrac{17}{2}\\\left(7,5\right)^2=y.\dfrac{17}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{32}{17}\left(cm\right)\\y=\dfrac{225}{34}\left(cm\right)\end{matrix}\right.\)

hình bên phải: 

Ta có: \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{6^2}+\dfrac{1}{8^2}=\dfrac{25}{276}\Rightarrow AH^2=\dfrac{276}{25}\)

\(\Rightarrow AH=\dfrac{24}{5}\Rightarrow z=\dfrac{24}{5}\)

Ta có: \(BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)

\(\Rightarrow y=10\left(cm\right)\)

Ta có: \(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=\dfrac{36}{10}\left(cm\right)\Rightarrow x=\dfrac{36}{10}\left(cm\right)\)

c) \(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=1+\dfrac{\sqrt{30}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=1+\dfrac{\sqrt{6}\left(\sqrt{5}-1\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}\)

\(=1+\dfrac{\sqrt{6}}{2\sqrt{2}}=1+\dfrac{\sqrt{3}}{2}=\dfrac{2+\sqrt{3}}{2}\)

d) \(\sqrt{\left(1-\sqrt{2016}\right)^2}.\sqrt{2017+2\sqrt{2016}}=\left|1-\sqrt{2016}\right|\sqrt{\left(\sqrt{2016}+1\right)^2}\)

\(=\left(\sqrt{2016}+1\right)\left(\sqrt{2016}-1\right)=2015\)

e) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=27+12\sqrt{5}+12\sqrt{5}=27+24\sqrt{5}\)

f) \(2\sqrt{5}\left(2-3\sqrt{5}\right)+\left(1-2\sqrt{5}\right)^2+6\sqrt{5}\)

\(=4\sqrt{5}-30+21-4\sqrt{5}+6\sqrt{5}=6\sqrt{5}-9\)

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

\(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{12}{4}=3\)

b) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\dfrac{-\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-3}{\sqrt{6}}=\dfrac{-\sqrt{3}}{\sqrt{2}}\)

c) \(\left(2+\sqrt{5}+\sqrt{3}\right)\left(2+\sqrt{5}-\sqrt{3}\right)=\left(2+\sqrt{5}\right)^2-3\)

\(=9+4\sqrt{5}-3=6+4\sqrt{5}\)

d) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

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

Xét \(x\ge1\Rightarrow\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

Xét \(0\le x< 1\Rightarrow\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\)

b) \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\sqrt{\dfrac{3\left(x+y\right)^2}{4}}\)

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

c) \(\dfrac{x+\sqrt{7}}{x^2+2x\sqrt{7}+7}\left(x\ne-\sqrt{7}\right)=\dfrac{x+\sqrt{7}}{\left(x+\sqrt{7}\right)^2}=\dfrac{1}{x+\sqrt{7}}\)

d) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(\dfrac{5\sqrt{7}+7\sqrt{5}}{\sqrt{35}}=\dfrac{\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right)}{\sqrt{35}}=\sqrt{7}+\sqrt{5}\)

\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}.6}-\sqrt{24.6}+\sqrt{\dfrac{50}{3}.6}\)

\(=\sqrt{16}-\sqrt{144}+\sqrt{100}=4-12+10=2\)

\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{2}\left(3-2\right)=\sqrt{2}\)

\(\sqrt{3x-5}=7\left(x\ge\dfrac{5}{3}\right)\Rightarrow3x-5=49\Rightarrow3x=54\Rightarrow x=18\)