ĐK:  \(x\ge0\)

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

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

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

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

Nhận thấy \(P\ge0\)nên P lớn nhất khi 1/P nhỏ nhất

=>  \(\frac{x-\sqrt{x}+1}{\sqrt{x}}\)nhỏ nhất

Ta có:  \(\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}-1=1\)

=> MIN 1/P = 1  khi x = 1

Vậy MAX P = 1 khi x = 1

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

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

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

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

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

\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac{\sqrt{x}}{\left(\sqrt{x}-1^2\right)+\sqrt{x}}\le1\)

\(Pmax=1\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow x=1\)

kb nha.

\(x\ge16\)

\(\Rightarrow\)\(\sqrt{x}\ge4\)

\(\Rightarrow\)\(\sqrt{x}-2\ge2\)

\(\Rightarrow\)\(\frac{1}{\sqrt{x}-2}\le\frac{1}{2}\)

\(\Rightarrow\)\(\frac{3}{\sqrt{x}-2}\le\frac{3}{2}\)

Dấu "=" xảy ra khi x = 16

Vậy MAX \(\frac{3}{\sqrt{x}-2}=\frac{3}{2}\)khi x = 16

\(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{2\sqrt{x}-9-\left(x-9\right)+2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x-3}}\)

sai r bạn ơi

b) \(< =>\sqrt{x+1}\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=0\)

<=>    x=-1

hoặc \(x^2-x+1=x+3\)    =>    \(x^2-2x-2=0...\)

Ta  có :\(sin^2a+cos^2a=1\)

Thay số: \(\left(\frac{2}{3}\right)^2\)\(+cos^2a=1\)\(\Rightarrow cos^2a=\frac{5}{9}\)

A=\(2sin^2a+5cos^2a\)\(\Rightarrow2.\frac{4}{9}+5.\frac{5}{9}\)\(\Rightarrow A=\frac{11}{3}\)

a, Ta có: \(BH//CF\left(gt\right)\)

               \(CF\perp AB\left(gt\right)\)

\(\Rightarrow BH\perp AB\)

\(\Delta ABH\)có: \(\widehat{ABH}=90^o,BH\perp AB\)

\(\Rightarrow AB^2=AE.AH\)(hệ thức lượng trong tam giác vuông)

\(AB=AC\left(gt\right)\)

\(\Rightarrow AC^2=AH.AE\)

Câu b chiều mình làm nhé

b, Kẻ \(DG\perp AB\)

Ta có: \(DG\perp AB\left(cd\right)\)

          \(FC\perp AB\left(gt\right)\)

\(\Rightarrow DG//FC\)

\(\Delta ABC\)cân tại A có: AD là đường cao của \(\Delta ABC\)\(\Rightarrow\)AD là đường trung tuyến của \(\Delta ABC\)\(\Rightarrow BD=DC\)

\(\Delta BEC\)có: \(DG//FC\left(cmt\right)\)

                       \(BD=DC\left(cmt\right)\)

\(\Rightarrow GF=FB\)

\(\Delta BFC\)có: \(GF=FB\left(cmt\right)\)

                        \(BD=DC\left(cmt\right)\)

\(\Rightarrow\)\(\hept{\begin{cases}DG//FC\\DG=\frac{1}{2}FC\end{cases}}\)

\(\Delta ADB\)có: \(\widehat{ADB}=90^o,DG\perp AB\)

\(\Rightarrow\frac{1}{DG^2}=\frac{1}{AD^2}+\frac{1}{DB^2}\)(hệ thức về cạnh và đường cao trong tam giác vuông)

mà \(DG=\frac{1}{2}FC\left(cmt\right)\)

\(\Rightarrow\frac{4}{FC^2}=\frac{1}{AD^2}+\frac{1}{DB^2}\)

mà \(BD=\frac{1}{2}BC\left(cmt\right)\)

\(\Rightarrow\frac{4}{FC^2}=\frac{1}{AD^2}+\frac{4}{BC^2}\)\(\Leftrightarrow\frac{1}{FC^2}=\frac{1}{4AD^2}+\frac{1}{BC^2}\)