Ta có: \(\frac58\left(x-2\right)\left(\frac{90}{x}+1\right)+\frac38\left(x-2\right)\cdot\frac{90}{x}=90\)

=>\(\frac58\cdot\left(90+x-\frac{180}{x}-2\right)+\frac{270}{8x}\left(x-2\right)=90\)

=>\(\frac{450}{8}+\frac58x-\frac{900}{8x}-\frac54+\frac{270}{8}-\frac{540}{8x}=90\)

=>\(\frac{720}{8}+\frac58x-\frac{1440}{8x}-\frac54-90=0\)

=>\(\frac58x-\frac{180}{x}-\frac54=0\)

=>\(\frac{5x-10}{8}-\frac{180}{x}=0\)

=>\(\frac{x\left(5x-10\right)-180\cdot8}{8x}=0\)

=>x(5x-10)-1440=0

=>5x(x-2)-1440=0

=>x(x-2)-288=0

=>\(x^2-2x-288=0\)

=>(x-18)(x+16)=0

=>x=18(nhận) hoặc x=-16(loại)

Khi x=18 thì y=90/18=5(nhận)

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

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

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

g: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

=>\(\frac{14\left(5x-5+2\right)}{84}-\frac{21\left(7x-1\right)}{84}=\frac{24\left(2x+1\right)}{84}-\frac{420}{84}\)

=>14(5x-3)-21(7x-1)=24(2x+1)-420

=>70x-42-147x+21=48x+24-420

=>-77x-21=48x-396

=>-125x=-396+21=-375

=>x=3

h: \(\frac{x-1}{2}+\frac{x-1}{4}=1-\frac{2\left(x-1\right)}{3}\)

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

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

=>9(x-1)=4(-2x+5)

=>9x-9=-8x+20

=>17x=29

=>x=29/17

i: \(\frac{x-3}{4}-\frac{5-2x}{6}=\frac{x}{2}+\frac{5}{12}\)

=>\(\frac{3\left(x-3\right)}{12}+\frac{2\left(2x-5\right)}{12}=\frac{6x}{12}+\frac{5}{12}\)

=>3(x-3)+2(2x-5)=6x+5

=>3x-9+4x-10=6x+5

=>7x-19=6x+5

=>x=24

j: \(\frac{x+2}{12}-\frac{x-3}{18}=\frac{x-5}{9}-\frac{x+4}{6}\)

=>\(\frac{3\left(x+2\right)-2\left(x-3\right)}{36}=\frac{4\left(x-5\right)-6\left(x+4\right)}{36}\)

=>3(x+2)-2(x-3)=4(x-5)-6(x+4)

=>3x+6-2x+6=4x-20-6x-24

=>x+12=-2x-44

=>3x=-56

=>x=-56/3

k: \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

=>\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

=>\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

=>x+10=0

=>x=-10

l: \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

=>\(\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+1-\frac{x}{2003}\)

=>\(\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

=>2003-x=0

=>x=2003

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=\(\left\{-4;0\right\}\)

k: ĐKXĐ: x<>5; x<>-5

\(\frac{7}{6x+30}+\frac{3}{4x-20}=\frac{15}{2x^2-50}\)

=>\(\frac{7}{6\left(x+5\right)}+\frac{3}{4\left(x-5\right)}=\frac{15}{2\left(x-5\right)\left(x+5\right)}\)

=>\(\frac{7\cdot2\cdot\left(x-5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{3\cdot3\cdot\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}=\frac{90}{12\left(x-5\right)\left(x+5\right)}\)

=>14(x-5)+9(x+5)=90

=>14x-70+9x+45=90

=>23x-25=90

=>23x=115

=>x=5(loại)

l: ĐKXĐ: x<>1; x<>-1

\(\frac{x}{x-1}+\frac{2x}{1-x^2}=0\)

=>\(\frac{x}{x-1}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x\left(x+1\right)-2x}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x\left(x+1-2\right)}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x}{x+1}=0\)

=>x=0(nhận)

m: ĐKXĐ: x<>2; x<>-2

\(\frac{x}{x+2}+\frac{5x+3}{x^2-4}=\frac{x}{x-2}\)

=>\(\frac{x\left(x-2\right)+5x+3}{\left(x-2\right)\left(x+2\right)}=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>x(x-2)+5x+3=x(x+2)

=>\(x^2-2x+5x+3=x^2+2x\)

=>3x+3=2x

=>3x-2x=-3

=>x=-3(nhận)

n: ĐKXĐ: x<>8

\(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac18=\frac{13x-102}{3x-24}\)

=>\(\frac{3\cdot12}{24\left(x-8\right)}+\frac{24\left(3x-20\right)}{24\left(x-8\right)}+\frac{3\left(x-8\right)}{24\left(x-8\right)}=\frac{8\left(13x-102\right)}{24\left(x-8\right)}\)

=>36+72x-480+3x-24=104x-816

=>104x-816=75x-468

=>29x=-468+816=348

=>x=12(nhận)

o: ĐKXĐ: x<>1/2; x<>-1/2

\(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}=\frac{8}{4x^2-1}\)

=>\(\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{8}{\left(2x-1\right)\left(2x+1\right)}\)

=>\(\left(2x+1\right)^2-\left(2x-1\right)^2=8\)

=>\(4x^2+4x+1-\left(4x^2-4x+1\right)=8\)

=>8x=8

=>x=1(nhận)

p:

ĐKXĐ: x<>-1; x<>2

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

=>\(\frac{x-2-5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{-15}{\left(x+1\right)\left(x-2\right)}\)

=>x-2-5(x+1)=-15

=>x-2-5x-5=-15

=>-4x-7=-15

=>4x+7=15

=>4x=8

=>x=2(loại)

q: ĐKXĐ: x<>1; x<>3

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

=>\(\frac{\left(x+5\right)\cdot\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

=>(x+5)(x-3)=(x+1)(x-1)-8

=>\(x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=6

=>x=3(loại)

r: ĐKXĐ: x<>1/3; x<>-1/3

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

=>\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

=>\(\left(3x-1\right)^2-\left(3x+1\right)^2=12\)

=>\(9x^2-6x+1-9x^2-6x-1=12\)

=>-12x=12

=>x=-1(nhận)

s: ĐKXĐ: x<>2; x<>-2

\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)

=>\(\frac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{12+x^2-4}{\left(x-2\right)\left(x+2\right)}\)

=>(x+1)(x+2)-5(x-2)=\(x^2+8\)

=>\(x^2+3x+2-5x+10=x^2+8\)

=>-2x+12=8

=>-2x=-4

=>x=2(loại)

k: ĐKXĐ: x<>5; x<>-5

\(\frac{7}{6x+30}+\frac{3}{4x-20}=\frac{15}{2x^2-50}\)

=>\(\frac{7}{6\left(x+5\right)}+\frac{3}{4\left(x-5\right)}=\frac{15}{2\left(x-5\right)\left(x+5\right)}\)

=>\(\frac{7\cdot2\cdot\left(x-5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{3\cdot3\cdot\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}=\frac{90}{12\left(x-5\right)\left(x+5\right)}\)

=>14(x-5)+9(x+5)=90

=>14x-70+9x+45=90

=>23x-25=90

=>23x=115

=>x=5(loại)

l: ĐKXĐ: x<>1; x<>-1

\(\frac{x}{x-1}+\frac{2x}{1-x^2}=0\)

=>\(\frac{x}{x-1}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x\left(x+1\right)-2x}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x\left(x+1-2\right)}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x}{x+1}=0\)

=>x=0(nhận)

m: ĐKXĐ: x<>2; x<>-2

\(\frac{x}{x+2}+\frac{5x+3}{x^2-4}=\frac{x}{x-2}\)

=>\(\frac{x\left(x-2\right)+5x+3}{\left(x-2\right)\left(x+2\right)}=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>x(x-2)+5x+3=x(x+2)

=>\(x^2-2x+5x+3=x^2+2x\)

=>3x+3=2x

=>3x-2x=-3

=>x=-3(nhận)

n: ĐKXĐ: x<>8

\(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac18=\frac{13x-102}{3x-24}\)

=>\(\frac{3\cdot12}{24\left(x-8\right)}+\frac{24\left(3x-20\right)}{24\left(x-8\right)}+\frac{3\left(x-8\right)}{24\left(x-8\right)}=\frac{8\left(13x-102\right)}{24\left(x-8\right)}\)

=>36+72x-480+3x-24=104x-816

=>104x-816=75x-468

=>29x=-468+816=348

=>x=12(nhận)

o: ĐKXĐ: x<>1/2; x<>-1/2

\(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}=\frac{8}{4x^2-1}\)

=>\(\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{8}{\left(2x-1\right)\left(2x+1\right)}\)

=>\(\left(2x+1\right)^2-\left(2x-1\right)^2=8\)

=>\(4x^2+4x+1-\left(4x^2-4x+1\right)=8\)

=>8x=8

=>x=1(nhận)

p:

ĐKXĐ: x<>-1; x<>2

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

=>\(\frac{x-2-5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{-15}{\left(x+1\right)\left(x-2\right)}\)

=>x-2-5(x+1)=-15

=>x-2-5x-5=-15

=>-4x-7=-15

=>4x+7=15

=>4x=8

=>x=2(loại)

q: ĐKXĐ: x<>1; x<>3

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

=>\(\frac{\left(x+5\right)\cdot\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

=>(x+5)(x-3)=(x+1)(x-1)-8

=>\(x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=6

=>x=3(loại)

r: ĐKXĐ: x<>1/3; x<>-1/3

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

=>\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

=>\(\left(3x-1\right)^2-\left(3x+1\right)^2=12\)

=>\(9x^2-6x+1-9x^2-6x-1=12\)

=>-12x=12

=>x=-1(nhận)

s: ĐKXĐ: x<>2; x<>-2

\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)

=>\(\frac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{12+x^2-4}{\left(x-2\right)\left(x+2\right)}\)

=>(x+1)(x+2)-5(x-2)=\(x^2+8\)

=>\(x^2+3x+2-5x+10=x^2+8\)

=>-2x+12=8

=>-2x=-4

=>x=2(loại)