Bài 5 hình 1: (tự vẽ hình nhé bạn)
a) Xét ΔABD và ΔACB ta có:
\(\widehat{BAD}\)= \(\widehat{BAC}\) (góc chung)
\(\widehat{ABD}\)= \(\widehat{ACB}\) (gt)
=> ΔABD ~ ΔACB (g-g)
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CB}\) = \(\dfrac{AD}{AB}\) (tsđd)
b) Ta có: \(\dfrac{AB}{AC}\) = \(\dfrac{AD}{AB}\) (cm a)
=> \(AB^2\) = AD.AC
=> \(2^2\) = AD.4
=> AD = 1 (cm)
Ta có: AC = AD + DC (D thuộc AC)
      => 4   =   1   + DC
      => DC = 3 (cm)
c) Xét ΔABH và ΔADE ta có: 
   \(\widehat{AHB}\) = \(\widehat{AED}\) (=\(90^0\))
   \(\widehat{ADB}\) = \(\widehat{ABH}\) (ΔABD ~ ΔACB)
=> ΔABH ~ ΔADE
=> \(\dfrac{AB}{AD}\) = \(\dfrac{AH}{AE}\) = \(\dfrac{BH}{DE}\) (tsdd)
Ta có: \(\dfrac{S_{ABH}}{S_{ADE}}\) = \(\left(\dfrac{AB}{AD}\right)^2\)= \(\left(\dfrac{2}{1}\right)^2\)= 4
=> đpcm

Tiếp bài 5 hình 2 (tự vẽ hình)
a) Xét ΔABC vuông tại A ta có:
\(BC^2\) = \(AB^2\) + \(AC^2\)
\(BC^2\) = \(21^2\) + \(28^2\)
BC = 35 (cm)
b) Xét ΔABC và ΔHBA ta có:
\(\widehat{BAC}\) = \(\widehat{AHB}\) ( =\(90^0\))
\(\widehat{ABC}\) = \(\widehat{ABH}\) (góc chung)
=> ΔABC ~ ΔHBA (g-g)
=> \(\dfrac{AB}{BH}\) = \(\dfrac{BC}{AB}\) (tsdd)
=> \(AB^2\) = BH.BC
=> \(21^2\) = 35.BH
=> BH = 12,6 (cm)
c) Xét ΔABC ta có:
BD là đường p/g (gt)
=> \(\dfrac{AD}{DC}\) = \(\dfrac{AB}{BC}\) (t/c đường p/g)
Xét ΔABH ta có: 
BE là đường p/g (gt)
=> \(\dfrac{HE}{AE}\) = \(\dfrac{BH}{AB}\) (t/c đường p/g)
Mà: \(\dfrac{AB}{BC}\) = \(\dfrac{BH}{AB}\) (cm b)
=> đpcm
d) Ta có: \(\left\{{}\begin{matrix}\widehat{HBE}+\widehat{BEH}=90^0\\\widehat{ABD}+\widehat{ADB=90^0}\\\widehat{HBE}=\widehat{ABD}\end{matrix}\right.\)
=> \(\widehat{BEH}=\widehat{ADB}\)
Mà \(\widehat{BEH}=\widehat{AED}\) (2 góc dd)
Nên \(\widehat{ADB}=\widehat{AED}\)
=> đpcm

1)x>=0

2)v8+5v2-v32+6v1/2=2v2+5v2-4v2+3v2=9v2

3)vx+1=2

x+1=4=>x=3

\(\left\{{}\begin{matrix}y-\dfrac{2}{5}=\dfrac{x}{50}\\y+1=\dfrac{x}{40}\end{matrix}\right.\)

`=> y -2/5 -y-1 = x/50 -x/40`

`<=> -7/5 = x(1/50-1/40)`

`=> x= -7/5 : (1/50 -1/40) `

`<=> x =280`

`=> y +1 =280/40 = 7`

`<=> y = 6`

Vậy.....

Bài 4: 

a: Thay x=36 vào A, ta được:

\(A=\dfrac{6+3}{6-2}=\dfrac{9}{4}\)

b: Ta có: \(B=\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{2}{x-4}\)

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

\(=\dfrac{x+3\sqrt{x}}{x-4}\)

a) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\) (1)

ĐKXĐ: x≠ 2

(1) ⇔ \(\dfrac{2x-10}{6\left(x-2\right)}-\dfrac{3x-6}{6\left(x-2\right)}=\dfrac{6x-9}{6\left(x-2\right)}\)

⇒ 2x - 10 - 3x + 6 = 6x - 9

⇔ -7x = -5

⇔ x = \(\dfrac{5}{7}\) (TMĐKXĐ)

Vậy S=\(\left\{\dfrac{5}{7}\right\}\)

b)\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\) (2)

ĐKXĐ: x≠ \(\pm2\)

(2) ⇒ x² - 3x +2 - x² - 2x = 5x - 2

⇔ -10x = -4

⇔ x = \(\dfrac{2}{5}\) (TMĐKXĐ)

Vậy S= \(\left\{\dfrac{2}{5}\right\}\)

c) \(\dfrac{6}{x+5}-\dfrac{1}{5-x}=\dfrac{3x+5}{x^2-25}\) (3)

ĐKXĐ: x ≠ \(\pm5\)

(3) ⇒ 6x - 30 -x +5 = 3x + 5

⇔ 2x = 30

⇔ x = 15 (TMĐKXĐ)

Vậy S= \(\left\{15\right\}\)

d) ĐKXĐ: x≠ \(\pm2\)

⇒ x² - 3x + 2 - x² - 2x = 2 - 5x

⇔ 0x = 0 (TMĐKXĐ)

Vậy PT có vô số nghiệm

e) ĐKXĐ: x≠ 0; x≠ 2

⇒ x² + 2x = x - 2 + 2 

⇔ x² + x = 0

⇔ x(x + 1) = 0

⇔\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy S = \(\left\{-1\right\}\)

f) ĐKXĐ: x≠ \(\pm2\)

⇒ x² - 2x + x + 2 = 2 - 3x

⇔ x² + 4x = 0

⇔ x(x+4) = 0

⇔\(\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-4\left(TMĐKXĐ\right)\end{matrix}\right.\)\(\)

Vậy S=\(\left\{-4;0\right\}\)