a) \(\frac{2x}{x+2}+\frac{x+2}{2x}=2\)

\(\Leftrightarrow4x^2+\left(x+2\right)^2=4x\left(x+2\right)\)

\(\Leftrightarrow5x^2+4x+4=4x^2+8x\)

\(\Leftrightarrow5x^2+4x+4-4x^2-8x=0\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow x^2-2.x.2+2^2=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Rightarrow x=2\)

\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)

Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)

khó quá mk mới học lớp 6 nên k giải đc thông cảm cho mk nha

có ai giúp mk giải 2 bài này vs 

quy dong tung phan thuc mot di ban

1) Hình như đề bị sai rồi bạn.

Thông thường pt đã cho sẽ là \(\frac{2x}{x-2}-\frac{5}{x-3}=\frac{5}{x^2-5x+6}\)

Ta thấy \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

Nên ĐKXĐ là \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

pt đã cho \(\Leftrightarrow\frac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{5\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{2x^2-6x-5x+10}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)\(\Rightarrow2x^2-11x+5=0\)(*)

Ta có \(\Delta=\left(-11\right)^2-4.2.5=81>0\)nên pt (*) có 2 nghiệm phân biệt:

\(\orbr{\begin{cases}x_1=\frac{-\left(-11\right)+\sqrt{81}}{2.2}=5\left(nhận\right)\\x_2=\frac{-\left(-11\right)-\sqrt{81}}{2.2}=\frac{1}{2}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{1}{2};5\right\}\)

2) Nhận thấy \(3x^2-27=3\left(x^2-9\right)=3\left(x-3\right)\left(x+3\right)\)nên ĐKXĐ ở đây là \(x\ne\pm3\)

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

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

\(\Leftrightarrow\frac{1-3x-9}{3x^2-27}=\frac{1}{4}\)\(\Rightarrow-12x-32=3x^2-27\)\(\Leftrightarrow3x^2+12x+5=0\)(#)

Nhận thấy \(\Delta'=6^2-3.5=21>0\)

Vậy pt (#) có 2 nghiệm phân biệt \(\orbr{\begin{cases}x_1=\frac{-12+\sqrt{21}}{3}\left(nhận\right)\\x_2=\frac{-12-\sqrt{21}}{3}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{-12\pm\sqrt{21}}{3}\right\}\)

Điều kiện xác định x khác 1

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

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

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

\(\Leftrightarrow x^2-3x^2-2x^2+x+2x+1=0\)

\(\Leftrightarrow-4x^2+2x+1=0\)

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

\(\Leftrightarrow-2x-1=0\)

\(\Rightarrow x=-0,5\)(thỏa mãn)

ok cám ơn bạn rất nhiều!

x+2x=3x mà bạn!

Bạn chỉ cần quy đồng khử mẫu thôi

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

\(\frac{6.\left(3x-1\right)-3x-1}{3}-\frac{12x+2\left(1-2x\right)}{2}=\frac{3\left(3x-1\right)}{5}\)

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

\(\Leftrightarrow\)\(\frac{15x-3}{3}-\frac{8x+2}{2}=\frac{9x-3}{5}\)

\(\Leftrightarrow\)\(10\left(15x-3\right)-15\left(8x+2\right)=6\left(9x-3\right)\)

\(\Leftrightarrow\)\(150x-30-120x-30=54x-18\)

\(\Leftrightarrow\)\(150x-120x-54x=-18+30+30\)

\(\Leftrightarrow-24x=42\)

\(\Leftrightarrow x=-\frac{7}{4}\)

<=>\(\frac{6x-2-x+1}{6}-\frac{6x+1-2x}{6}=\frac{3x-1}{10}\)

<=>\(\frac{5\left(x-2\right)}{30}=\frac{3\left(3x-1\right)}{30}\)

<=> 5x - 2 = 9x - 3

<=> 4x = 1

=> x = 1/4

Vậy S = {1/4}

a , 2x -3 = 5x + 6

    2x -5x=6+3

    -3x = 9

     x =9 :(-3)

   x= -3

a) 2x-5x=3+6

-3x=9

x=-3

vậy........

b)(2x+1).(3x-2)-(5x-8).(2x+1)=0

(2x+1).(3x-2-2x-1)=0

(2x-1).(x-3)=0

==>x=1/2 ; x=3

c)(2x+1).5-(7x+5)=(2x-2).3

10x+5-7x-5=6x-6

3x=6x-6

3x-6x=6

-3x=6

x=-2

a) 2x - 3 = 5x + 6

<=> -3x = 9

<=> x = -3

b) (2x + 1).(3x - 2) = (5x - 8).(2x + 1)

<=> 6x² - 4x + 3x - 2 = 10x² + 5x - 16x -8

<=> -4x² - 10x + 6 = 0

<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

c) \(\frac{2\text{x}+1}{3}-\frac{7\text{x}+5}{15}=\frac{2\text{x}-2}{5}\)

<=> \(\frac{5.\left(2\text{x}+1\right)}{5.3}-\frac{7\text{x}+5}{15}=\frac{3.\left(2\text{x}-2\right)}{3.5}\)

<=> 10x + 5 - 7x + 5 - 6x + 6 = 0

<=> -3x + 16 = 0

<=> -3x = -16

<=> x = \(\frac{16}{3}\)

d) \(\frac{3x}{x-2}-\frac{x}{x-5}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> \(\frac{3\text{x}\left(x-5\right)-x\left(x-2\right)}{\left(x-2\right).\left(5-x\right)}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> 3x² - 15x - x² + 2x - 3x = 0

*Câu e dễ quá, bạn tự làm nhé :v*