vì ta có 12-3x/32 =12-3 nhân x 

  mà rút gon lại bằng 6 ,còn 32 rút gọn lại bằng 4 nên x=0

VD:    12-3 nhân 0 /32

        =12-0 / 32

       =12 / 32 và rút gọn bằng 6/4

 nên x=0

hok tốt

\(\frac{12-3x}{32}=\frac{6}{4-x}ĐK:x\ne4\)

\(\Leftrightarrow\frac{\left(12-3x\right)\left(4-x\right)}{32\left(4-x\right)}=\frac{192}{32\left(4-x\right)}\)

\(\Leftrightarrow48-12x-12x+3x^2=192\)

\(\Leftrightarrow48-24x+3x^2=192\)Xử nốt nhé, dễ rồi!!! 

\(\frac{12-3x}{32}=\frac{6}{4-x}\)

\(\Leftrightarrow\left(12-3x\right)\left(4-x\right)=32.6\)

\(\Leftrightarrow48-12x-12x+3x^2=192\)

\(\Leftrightarrow48-24x+3x^2=192\)

\(\Leftrightarrow3\left(18-8x-x^2\right)=192\)

\(\Leftrightarrow\left(4-x\right)^2=64\)

\(\Leftrightarrow\left(4-x\right)^2=8^2\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}\)

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

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

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

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

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

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

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

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

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

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

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

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

a) \(\frac{36\left(x-2\right)}{32-16x}=\frac{36\left(x-2\right)}{16\left(2-x\right)}=-\frac{36\left(2-x\right)}{16\left(2-x\right)}=-\frac{36}{16}=-\frac{9}{4}\)

b) \(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}=\frac{3x-6}{x^3+2x^2+4x}\)

c) \(\frac{7x^2+14x+7}{3x^2+3x}=\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}=\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\frac{7\left(x+1\right)}{3x}=\frac{7x+7}{3x}\)

d) \(\frac{x^4-5x^2+4}{x^4-10x^2+9}=\frac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}=\frac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}=\frac{\left(x^2-4\right)\left(x^2-1\right)}{\left(x^2-9\right)\left(x^2-1\right)}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\)

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

\(ĐKXĐ:x\ne\pm2\)

\(\frac{x}{x+2}+\frac{6}{2-x}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow\frac{x}{x+2}-\frac{6}{x-2}-\frac{3x-12}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)-6\left(x+2\right)-\left(3x-12\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2-2x-6x-12-3x+12=0\)

\(\Leftrightarrow x^2-11x=0\)

\(\Leftrightarrow x\left(x-11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-11=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=11\end{cases}}\)(tm)

Vậy tập nghiệm của phương trình là \(S=\left\{0;11\right\}\)

\(ĐKXĐ:x\ne\pm2\)

\(\frac{x}{x+2}+\frac{6}{2-x}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow\frac{x}{x+2}+\frac{-6}{x-2}-\frac{3x-12}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{-6\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3x-12}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)-6\left(x+2\right)-\left(3x-12\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x\left(x-2\right)-6\left(x+2\right)-\left(3x-12\right)=0\)

\(\Leftrightarrow x^2-2x-6x-12-3x+12=0\)

\(\Leftrightarrow x^2-11x=0\)\(\Leftrightarrow x\left(x-11\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=11\end{cases}}\)( thoả mãn \(ĐKXĐ\))

Vậy tập nghiệm của phương trình là \(S=\left\{0;11\right\}\)