mình nghĩ đề là ABC vuông tại A nhé, vì mình thử đủ mọi cách r ;))) 

Xét tam giác ABC vuông tại A, đường cao AH

cosB = \(\frac{AB}{BC}=\frac{3}{5}\Rightarrow AB=\frac{3}{5}.30=18\)cm 

cosB = \(\frac{BH}{AB}=\frac{3}{5}\Rightarrow BH=\frac{3}{5}.18=\frac{54}{5}\)cm 

Ta có : cosB = \(\frac{BH}{AB}=\frac{3}{5}\Rightarrow\frac{BH}{3}=\frac{AB}{5}\Rightarrow\frac{BH^2}{9}=\frac{AB^2}{25}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{AB^2}{25}=\frac{BH^2}{9}=\frac{AB^2-BH^2}{16}=\frac{AH^2}{16}\)

\(\Rightarrow\frac{AB^2}{25}=\frac{AH^2}{16}\Rightarrow AH^2=\frac{18^2.16}{25}=\frac{5184}{25}\Rightarrow AH=\frac{72}{5}\)cm 

ta có 

\(A=111..1000..0+222..2+3=10^{2007}\left(1+10+..+10^{2004}\right)+2.\left(1+10+..+10^{2006}\right)+3\)

\(=10^{2007}.\frac{10^{2005}-1}{9}+2.\frac{10^{2007}-1}{9}+3=\frac{10^{2.2006}-10.10^{2006}+25}{9}=\left(\frac{10^{2006}-5}{3}\right)^2\)

rõ ràng Alà số tự nhiên nên \(\left(\frac{10^{2006}-5}{3}\right)\) là số tự nhiên, vậy ta có đpcm

a, Thay x = 49 ta được : \(M=\frac{49+7}{7-5}=\frac{56}{2}=28\)

b, Với \(x\ge0;x\ne25\)

\(N=\frac{\sqrt{x}}{\sqrt{x}+5}+\frac{2\sqrt{x}-3}{\sqrt{x}-5}-\frac{2x-3\sqrt{x}-15}{x-25}\)

\(=\frac{x-5\sqrt{x}+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+5\right)-2x+3\sqrt{x}+15}{x-25}\)

\(=\frac{-x-2\sqrt{x}+15+2x+7\sqrt{x}-15}{x-25}=\frac{x+5\sqrt{x}}{x-25}=\frac{\sqrt{x}}{\sqrt{x}-5}\)

c, Ta có : \(P=\frac{M}{N}\Rightarrow\frac{x+7}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{x+7}{\sqrt{x}}=\sqrt{x}+\frac{7}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{7}{\sqrt{x}}}=2\sqrt{7}\)

Dấu ''='' xảy ra khi \(\sqrt{x}=\frac{7}{\sqrt{x}}\Rightarrow x=7\)riêng ý c thì mình ko chắc :(( 

a.\(x=49\Rightarrow M=\frac{49+7}{\sqrt{49}-5}=\frac{56}{2}=28\)

\(N=\frac{\sqrt{x}\left(\sqrt{x}-5\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+5\right)-2x+3\sqrt{x}+15}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-5\sqrt{x}+2x+7\sqrt{x}-15-2x+3\sqrt{x}+15}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{x+5\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)

\(c.P=M:N=\frac{x+7}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{x+7}{\sqrt{x}}=\sqrt{x}+\frac{7}{\sqrt{x}}\ge2\sqrt{7}\)

a/ \(\sqrt{2-x}-\sqrt{x^2-4}=0\)

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

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

Làm nốt

b/ \(\sqrt{x-3}-2\sqrt{x^2-9}=0\)

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

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

Làm nốt

a) ĐK : x >= 5

\(\Leftrightarrow\sqrt{x-5}+2\sqrt{25\left(x-5\right)}=22\)

\(\Leftrightarrow\sqrt{x-5}+10\sqrt{x-5}=22\)

\(\Leftrightarrow11\sqrt{x-5}=22\)

\(\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

b) ĐK : x >- -1/2

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

\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\Leftrightarrow\sqrt{2x+1}=3\Leftrightarrow2x+1=9\Leftrightarrow x=4\left(tm\right)\)

Lời giải:
Áp dụng hệ thức lượng trong tam giác vuông đối với tam giác ABC vuông, đường cao AH ta có:

\(AB^2=BH\cdot BC\\ AC^2=CH\cdot BC\\ \Rightarrow\frac{AB^2}{AC^2}=\frac{BH\cdot BC}{CH\cdot BC}=\frac{HB}{HC}\)

\(\Rightarrow\frac{HB}{HC}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)