Bài 1:

1, \(\frac{2x-5}{x+5}=3\) (ĐKXĐ: x \(\ne\) -5)

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

\(\Rightarrow\) 2x - 5 = 3(x + 5)

\(\Leftrightarrow\) 2x - 5 = 3x + 15

\(\Leftrightarrow\) 2x - 3x = 15 + 5

\(\Leftrightarrow\) -x = 20

\(\Leftrightarrow\) x = -20 (TMĐKXĐ)

Vậy S = {-20}

2, \(\frac{4}{x+1}=\frac{3}{x-2}\) (ĐKXĐ: x \(\ne\) -1; x \(\ne\) 2)

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

\(\Rightarrow\) 4(x - 2) = 3(x + 1)

\(\Leftrightarrow\) 4x - 8 = 3x + 3

\(\Leftrightarrow\) 4x - 3x = 3 + 8

\(\Leftrightarrow\) x = 11 (TMĐKXĐ)

Vậy S = {11}

3, \(\frac{5}{2x-3}=\frac{1}{x-4}\) (ĐKXĐ: x \(\ne\) \(\frac{3}{2}\); x \(\ne\) 4)

\(\Leftrightarrow\) \(\frac{5\left(x-4\right)}{\left(2x-3\right)\left(x-4\right)}=\frac{2x-3}{\left(2x-3\right)\left(x-4\right)}\)

\(\Rightarrow\) 5(x - 4) = 2x - 3

\(\Leftrightarrow\) 5x - 20 = 2x - 3

\(\Leftrightarrow\) 5x - 2x = -3 + 20

\(\Leftrightarrow\) 3x = 17

\(\Leftrightarrow\) x = \(\frac{17}{3}\) (TMĐKXĐ)

Vậy S = {\(\frac{17}{3}\)}

Bài 2:

1, \(\frac{1}{x-1}+\frac{2}{x+1}=\frac{5x-3}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)

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

\(\Rightarrow\) x + 1 + 2(x - 1) = 5x - 3

\(\Leftrightarrow\) x + 1 + 2x - 2 = 5x - 3

\(\Leftrightarrow\) 3x - 1 = 5x - 3

\(\Leftrightarrow\) 3x - 5x = -3 + 1

\(\Leftrightarrow\) -2x = -2

\(\Leftrightarrow\) x = 1 (KTM)

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

2, \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\) (ĐKXĐ: x \(\ne\) 2; x \(\ne\) 0)

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

\(\Rightarrow\) x(x + 2) - x + 2 = 2

\(\Leftrightarrow\) x² + 2x - x + 2 = 2

\(\Leftrightarrow\) x² + x = 2 - 2

\(\Leftrightarrow\) x² + x = 0

\(\Leftrightarrow\) x(x + 1) = 0

\(\Leftrightarrow\) x = 0 hoặc x + 1 = 0

\(\Leftrightarrow\) x = 0 và x = -1

Ta có: x = 0 KTM đkxđ

\(\Rightarrow\) x = -1

Vậy S = {-1}

3, \(\frac{5}{x-3}-\frac{3}{x+3}=\frac{3x}{x^2-9}\) (ĐKXĐ: x \(\ne\) \(\pm\) 3)

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

\(\Rightarrow\) 5(x + 3) - 3(x - 3) = 3x

\(\Leftrightarrow\) 5x + 15 - 3x + 9 = 3x

\(\Leftrightarrow\) 2x + 24 = 3x

\(\Leftrightarrow\) 2x - 3x = 24

\(\Leftrightarrow\) -x = 24

\(\Leftrightarrow\) x = -24 (TMĐKXĐ)

Vậy S = {-24}

Chúc bn học tốt!!

Mình tính mãi vẫn có chỗ sai, mong bạn thông cảm!!

a, ĐKXĐ \(\hept{\begin{cases}x\ne1\\x\ne2\\x\ne3\end{cases}x\ne4}\)

ta có \(đề\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)

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

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

              \(\Leftrightarrow\frac{1}{x-1}-\frac{2}{x-2}=\frac{3}{x-3}-\frac{4}{x-4}\)

             \(\Leftrightarrow\frac{x-2-2x+2}{\left(x-1\right)\left(x-2\right)}=\frac{3x-12-4x+12}{\left(x-3\right)\left(x-4\right)}\)

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

               \(\Leftrightarrow\left(x-1\right)\left(x-2\right)=\left(x-3\right)\left(x-4\right)\)(đến đây bạn nhân ra tự giải nhé )

p/s :mình nghĩ bạn viết sai đề đấu + ở phép đầu tiên ko phải - bạn xem lại nhé

b,\(\Leftrightarrow[2\left(x-3\right)]^2-\left(x-1\right)^2=0\)

  \(\Leftrightarrow\left(2x-6+x-1\right)\left(2x-6-x+1\right)=0\)

   \(\Leftrightarrow\left(3x-7\right)\left(x-5\right)=0\)(bạn tự giải)

c,\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

   \(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)

  \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\left(do\left(x^2+1>0\right)\right)\)

Cảm ơn bạn nhé

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

các bạn giúp mình với. cảm ơn 

giúp mình với

a,\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)\(\Leftrightarrow\frac{13\left(x+3\right)}{\left(x^2-9\right)\left(2x+7\right)}+\frac{x^2-9}{\left(x^2-9\right)\left(2x+7\right)}-\frac{6\left(2x+7\right)}{\left(x^2-9\right)\left(2x+7\right)}=0\)

\(\Leftrightarrow x+x^2-12=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=3\end{cases}}\)

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

\(\Leftrightarrow x^2-3x-10=0\Rightarrow\orbr{\begin{cases}x=5\left(L\right)\\x=-2\end{cases}}\)

c,\(\frac{1}{x+2}+\frac{1}{x\left(x-2\right)}-\frac{8}{x\left(x^2-4\right)}=0\)\(\Leftrightarrow\frac{x\left(x-2\right)}{x\left(x^2-4\right)}+\frac{x+2}{x\left(x^2-4\right)}-\frac{8}{x\left(x^2-4\right)}=0\)

\(\Leftrightarrow x^2-x-6=0\Rightarrow\orbr{\begin{cases}x=3\\x=-2\left(L\right)\end{cases}}\)

d,\(\frac{2}{\left(x^2-4\right)}-\frac{1}{x\left(x-2\right)}-\frac{x+4}{x\left(x+2\right)}=0\)\(\Leftrightarrow\frac{2x}{x\left(x^2-4\right)}-\frac{x+2}{x\left(x^2-4\right)}-\frac{\left(x+4\right)\left(x-2\right)}{x\left(x^2-4\right)}=0\)

\(\Leftrightarrow-x^2-5x-10=0\)(vô nghiệm)

\(\)

\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)

Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)

\(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\)

\(\Leftrightarrow4x-2-6x-3=4\)

\(\Leftrightarrow-2x=9\)

\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)

Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)

\(b,ĐKXĐ:x\ne\pm1;-3\)

Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)

\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)

\(\Leftrightarrow9x^2+14x+13=0\)

\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)

\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)

Pt vô nghiệm 

\(c,ĐKXĐ:x\ne1\)

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

\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)

\(\Leftrightarrow3x^2=3\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow x=\pm1\)

Kết hợp vs ĐKXĐ được x = -1

Vậy pt có nghiệm duy nhất x = -1

làm lần lượt nha(bài nào k bt bỏ qua)

\(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)

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

\(\Rightarrow-2x-5=4\)

\(\Rightarrow-2x=9\)

\(\Rightarrow x=\frac{9}{-2}\)

Chọn D.

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

\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)