\(\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne4\right)\\ =\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-2-2\sqrt{x}-4-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =-\dfrac{2}{\sqrt{x}-2}\)

dk là x khác 4 mới đúng nhee 

\(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\left(ĐKXĐ:x\ge0\right)\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)}+\dfrac{10-x}{\sqrt{x}+2}\)

\(=\dfrac{x-4+10-x}{\sqrt{x}+2}\)

\(=\dfrac{6}{\sqrt{x}+2}\)

\(=\dfrac{6\left(\sqrt{x}-2\right)}{x-4}\)

\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}+1\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\=\dfrac{\sqrt{x}+1-\sqrt{x}+1+x+\sqrt{x}-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\\=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\=1\)

\(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\)

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

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

\(\Leftrightarrow\dfrac{2}{3-1}\)

\(\Leftrightarrow1\)

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow\dfrac{1}{9}=\dfrac{1}{AB^2}+\dfrac{1}{25}\)

\(\Leftrightarrow\dfrac{1}{AB^2}=\dfrac{16}{225}\)

\(\Leftrightarrow AB=\dfrac{15}{4}\)

- \(AH.BC=AB.AC\)

\(3.BC=\dfrac{15}{4}.5\)

\(BC=6,25\)

- \(CH=\dfrac{AC^2}{BC}=4\)

=> BH = 6,25 - 4 = 2,25 

a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)

A C B H E F J I O

O là giao của AH và EF

\(AF\perp AB;HE\perp AB\) => AF//HE

\(AE\perp AC;HF\perp AC\) => AE//HF

=> AEHF là hình bình hành mà \(\widehat{A}=90^o\) => AEHF là HCN

\(\Rightarrow AH=EF\) (trong HCN hai đường chéo băng nhau)

\(OA=OH;OE=OF\) (trong hbh hai đường chéo cắt nhau tại trung điểm mỗi đường)

=> OE=OH => tg OEH cân tại O

Vì AEHF là HCN nên

\(\widehat{EAF}=\widehat{EHF}=90^o\) => A và H cùng nhìn EF dưới 1 góc vuông => AEHF là tứ giác nội tiếp đường tròn tâm O bán kính EF

Xét tg vuông BEH có

IB=IH (gt) \(\Rightarrow IE=IB=IH=\dfrac{BH}{2}\) (trong tg vuông trung tuyến thuộc cạnh huyền thì bằng nửa cạnh huyền)

=> tg IEH cân tại I \(\Rightarrow\widehat{IEH}=\widehat{IHE}\) (1)

tg OEH cân tại O (cmt) \(\Rightarrow\widehat{OEH}=\widehat{OHE}\) (2)

Mà \(\widehat{IHE}+\widehat{OHE}=\widehat{AHB}=90^o\) (3)

Từ (1) (2) (3) \(\Rightarrow\widehat{IEH}+\widehat{OEH}=\widehat{FEI}=90^o\)

\(\Rightarrow IE\perp EF\) mà EF là đường kính (O) => IE là tiếp tuyến đường tròn (O).

C/m tương tự ta cũng có \(JF\perp EF\) => JF cũng là tiếp tuyến với (O)

=> IE//JF (cùng vuông góc với EF)

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

\(x=36\Rightarrow P=\dfrac{\sqrt{36}+1}{\sqrt{36}-2}=\dfrac{6+1}{6-2}=\dfrac{7}{4}\)

\(x=6-2\sqrt{5}=\sqrt{5^2}-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)

                   \(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}-2}=\dfrac{\left|\sqrt{5}-1\right|+1}{\left|\sqrt{5}-1\right|-2}=\dfrac{\sqrt{5}}{\sqrt{5}-3}\)

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

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

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