\(A=\dfrac{x^2+6x+9-2x^2+6+x^2-3x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{2\left(x-3\right)\left(x+3\right)}{6\left(x-2\right)}\)

\(=\dfrac{3x+15}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{3\left(x-2\right)}=\dfrac{x+5}{x-2}\)

a) Đặt \(x-1=a\)

\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)

\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)

\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)

Vậy pt vô nghiệm

a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)

\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)

\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)

\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)

\(\frac{31}{2}=2\)

=> không có x thỏa mãn đề bài.

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

\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)

\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)

\(7-4x-3x^2=25x-25\)

\(7-4x-3x^2-25x+25=0\)

\(32-29x-3x^2=0\)

\(3x^2+29x-30=0\)

\(3x^2+32x-3x-32=0\)

\(x\left(3x+32\right)-\left(3x+32\right)=0\)

\(\left(3x+32\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)

lên mạng chứ 0 tự làm ak

\(\left(\frac{x}{x^2-36}+\frac{6-x}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)

đkxđ: \(x\ne0;x\ne\pm6\)

MTC=x(x+6)(x-6)

\(=\left[\frac{x}{\left(x+6\right)\left(x-6\right)}+\frac{6-x}{x\left(x+6\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)

\(=\left[\frac{x^2}{x\left(x^2-36\right)}-\frac{\left(x-6\right)^2}{x\left(x^2-36\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)

\(=\frac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)

\(=\frac{12}{x\left(x-6\right)}-\frac{x^2}{x\left(x-6\right)}\)

\(=\frac{12-x^2}{x\left(x-6\right)}\)

.....................

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

A = \(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)

= \(\left[\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)

= \(\left[\frac{x^2}{x\left(x-6\right)\left(x+6\right)}-\frac{\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)

= \(\frac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)

= \(\frac{\left(x-x+6\right)\left(x+x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)

=

= \(\frac{x\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2x-6}{x\left(x+6\right)}-\frac{x}{x-6}\)

= \(\frac{2x-6}{\left(x-6\right)\left(x+6\right)}.\frac{x\left(x+6\right)}{2x-6}\) \(-\frac{x}{x-6}\)

= \(\frac{x}{x-6}-\frac{x}{x-6}\)

= 0

\(C=\frac{\left(x+3\right)^2-2x^2+6+x\left(x-3\right)}{x^2-9}.\frac{2x^2-18}{6x-12}\)\(\)

\(C=\frac{x^2+6x+9-2x^2+6+x^2-3x}{x^2-9}.\frac{2\left(x^2-9\right)}{6x-12}\)\(C=\frac{3x+15}{6x-12}.2=\frac{x+5}{x-2}=1+\frac{7}{x-2}\)

Để C nguyên =>(x-2) thuộc Ư(7) \(\Rightarrow x\in\left\{3;1;9;-5\right\}\)

rzddddddfg