Đặt \(\hept{\begin{cases}\sqrt{x}=a\left(\ge0\right)\\M=\frac{4a}{a^2-4a+11}\end{cases}}\) \(\left(x\ge0\right)\)

Khi đó: \(a^2M-4aM+11M=4a\)

\(\Leftrightarrow a^2M-4a\left(M+1\right)+11M=0\)

\(\Delta^'=\left[-2\left(M+1\right)\right]^2-M\cdot11M\)

\(=4\left(M^2+2M+1\right)-11M^2=-7M^2+8M+4\)

\(\Rightarrow7M^2-8M-4\le0\Rightarrow\frac{4+2\sqrt{11}}{7}\ge M\ge\frac{4-2\sqrt{11}}{7}\)

Mà \(\hept{\begin{cases}4\sqrt{x}\ge0\\x-4\sqrt{x}+11>0\end{cases}\left(\forall xtm\right)}\)

\(\Rightarrow0\le M\le\frac{4+2\sqrt{11}}{7}\Rightarrow\hept{\begin{cases}M_{max}=\frac{4+2\sqrt{11}}{7}\\M_{min}=0\end{cases}}\)

Theo bài ra ta có 

AB + AH + BH = 30 

AC + CH + AH = 40

AB + BC + AC = 50 

Khi đó AB + AH + BH + AC + CH + AH = 70 

=> AB + AC + (BH + CH) + 2AH = 70

=> AB  + AC + BC + 2AH = 70

=> 50 + 2AH = 70

=> AH = 10

Vậy AH = 10 cm

Vì AB < AC nên trên cạnh AC lấy điểm F sao cho AB = AF

=> Tam giác ABF cân tại A

Ta có: AD = AE => BD = FE => BDEF là hình thang cân => BE = FD

Xét: Tam giác ABF cân tại A, ta có: AFB là góc ở đáy nên là góc nhọn

=> \(\widehat{AFD\:}\)là góc nhọn

=> \(\widehat{DFC}\)là góc tù

Vậy: CD > FD = BE

e cần gấp câu b,c,c ạ

\(\frac{-5\sqrt{x}+4}{3\sqrt{x}-2}+\frac{6\sqrt{x}+4}{2\sqrt{x}+3}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)

\(\frac{\left(-5\sqrt{x}+4\right)\left(2\sqrt{x}+3\right)+\left(6\sqrt{x}+4\right)\left(3\sqrt{x}-2\right)}{6x+5\sqrt{x}-6}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)

\(\frac{-10x-7\sqrt{x}+12+18x-8}{6x+5\sqrt{x}-6}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)

\(\frac{8x-7\sqrt{x}+4}{6x+5\sqrt{x}-6}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)

\(\frac{24x-21\sqrt{x}+12+29\sqrt{x}-28}{18x+15\sqrt{x}-18}\)

\(\frac{24x+8\sqrt{x}-16}{18x+15\sqrt{x}-18}\)

\(\frac{24x+24\sqrt{x}-16\sqrt{x}-16}{18x-12\sqrt{x}+27\sqrt{x}-18}\)

\(\frac{24\sqrt{x}\left(\sqrt{x}+1\right)-16\left(\sqrt{x}+1\right)}{6\sqrt{x}\left(3\sqrt{x}-2\right)+9\left(3\sqrt{x}-2\right)}\)

\(\frac{\left(\sqrt{x}+1\right)\left(24\sqrt{x}-16\right)}{\left(3\sqrt{x}-2\right)\left(6\sqrt{x}+9\right)}\)

\(\frac{\left(\sqrt{x}+1\right).8\left(3\sqrt{x}-2\right)}{\left(3\sqrt{x}-2\right)\left(6\sqrt{x}+9\right)}\)

\(\frac{8\sqrt{x}+8}{6\sqrt{x}+9}\)

đk: \(x\ge1\)

\(PT\Leftrightarrow\left(\sqrt{x-1}-2\right)+\left(\sqrt{2x-1}-3\right)=0\)

\(\Leftrightarrow\frac{x-1-4}{\sqrt{x-1}+2}+\frac{2x-1-9}{\sqrt{2x-1}+3}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt{x-1}+2}+\frac{2}{\sqrt{2x-1}+3}\right)=0\)

\(\Rightarrow x-5=0\Rightarrow x=5\left(tm\right)\)

Vậy x = 5

\(ĐK:x\ge1\)

\(\Leftrightarrow\left(\sqrt{x-1}-2\right)+\left(\sqrt{2x-1}-3\right)=0\)

\(\Leftrightarrow\frac{x-1-4}{\sqrt{x-1}+2}+\frac{2x-1-9}{\sqrt{2x-1}+3}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt{x-1}+2}+\frac{2}{\sqrt{2x-1}+3}\right)=0\)(*)

Dễ thấy với mọi x thỏa đk thì \(\frac{1}{\sqrt{x-1}+2}+\frac{2}{\sqrt{2x-1}+3}>0\)

nên (*) <=> x - 5 = 0 <=> x = 5 (tm)

Vậy pt có nghiệm duy nhất x = 5

\(\frac{x^2-3y}{x\left(1-3y\right)}=\frac{y^2-3x}{y\left(1-3x\right)}\)

\(\Rightarrow\left(x^2-3y\right)\left(y-3xy\right)=\left(y^2-3x\right)\left(x-3xy\right)\)

\(\Leftrightarrow x^2y-3x^3y-3y^2+9xy^2=xy^2-3xy^3-3x^2+9x^2y\)

\(\Leftrightarrow-3xy\left(x+y\right)\left(x-y\right)+3\left(x+y\right)\left(x-y\right)-8xy\left(x-y\right)=0\)

\(\Leftrightarrow3\left(x+y\right)-3xy\left(x+y\right)-8xy=0\)(vì \(x\ne y\)) 

\(\Leftrightarrow\frac{x+y}{xy}=x+y+\frac{8}{3}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=x+y+\frac{8}{3}\)

14, \(\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\)

\(=\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}\)

\(=\frac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)+\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(3\sqrt{x}+14\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}+14}{5\sqrt{x}-1}\)

thank

\(\frac{3\sqrt{2}-6}{\sqrt{2}-1}=\frac{3\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=3\)