x-3/-3  =  -27/x-3

=> (x-3)(x-3)=(-3)(-27)

=> (x-3)^2 = 81=9^2

=> x-3=9 hoặc x-3=-9

=> x=12 hoặc x=-6

Ta có :\(\frac{x-3}{-3}=\frac{-27}{x-3}\)

\(\Rightarrow\left(x-3\right)\left(x-3\right)=\left(-3\right).\left(-27\right)\) ( Tính chất tỉ lệ thức ) 

\(\Rightarrow\left(x-3\right)^2=81\)

\(\Rightarrow\left(x-3\right)^2=\left(\pm9\right)^2\)

\(\Rightarrow x-3=\pm9\)

\(\Rightarrow x=-6;12\)

làm luông di

đây nhá , chị nêu phương pháp cho 

\(\frac{A}{B}\)= \(\frac{C}{D}\) 

<=> A*D=B*C 

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(\Rightarrow x=x\)hoặc \(x=-x\)
---Nếu x = x
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
       \(x+\frac{1}{2}+x+\frac{2}{3}+x+\frac{3}{4}=4x\)
       \(3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
       \(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x-3x\)
         \(\Rightarrow x=\frac{23}{12}\)
---Nếu x = -x

\(\Rightarrow\left|-x+\frac{1}{2}\right|+\left|-x+\frac{2}{3}\right|+\left|-x+\frac{3}{4}\right|=4x\)
       \(-x+\frac{1}{2}+\left(-x\right)+\frac{2}{3}+\left(-x\right)+\frac{3}{4}=4x\)
        \(-3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
          \(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x+3x\)
           \(\Rightarrow7x=\frac{23}{12}\)
            \(\Rightarrow x=\frac{23}{12}:7\)
            \(\Rightarrow x=\frac{23}{12}.\frac{1}{7}\)
            \(\Rightarrow x=\frac{23}{84}\)
\(Vậyx=\frac{23}{12};x=\frac{23}{84}\)

\(a,2,5\cdot x=\frac{11}{15}\)

\(\Rightarrow x=\frac{11}{15}:2,5\)

\(\Rightarrow x=\frac{11}{15}:\frac{25}{10}\)

\(\Rightarrow x=\frac{11}{15}\cdot\frac{10}{25}\)

\(\Rightarrow x=\frac{11}{3}\cdot\frac{2}{25}=\frac{22}{75}\)

\(b,x-15\%\cdot x=\frac{1}{3}\)

\(\Rightarrow x-\frac{15}{100}\cdot x=\frac{1}{3}\)

\(\Rightarrow x-\frac{3}{20}\cdot x=\frac{1}{3}\)

\(\Rightarrow\frac{20x}{20}-\frac{3x}{20}=\frac{1}{3}\)

\(\Rightarrow\frac{17x}{20}=\frac{1}{3}\)

\(\Rightarrow17x\cdot3=20\)

\(\Rightarrow17x=\frac{20}{3}\)

\(\Rightarrow x=\frac{20}{3}:17=\frac{20}{3}\cdot\frac{1}{17}=\frac{20}{51}\)

Câu c mk làm sau :v

c,\(\left[\frac{3x}{7}+1\right]:\left[-4\frac{1}{7}\right]=\frac{-3}{28}\)

\(\Rightarrow\left[\frac{3x}{7}+1\right]:\left[-\frac{29}{7}\right]=\frac{-3}{28}\)

\(\Rightarrow\frac{3x}{7}+1=-\frac{3}{28}\cdot-\frac{29}{7}=\frac{87}{196}\)

\(\Rightarrow\frac{3x}{7}=\frac{87}{196}-1\)

Tìm nốt :v

Nhanh bạn ơi

Bài 1:

\(4.\left(\frac{-1}{2}\right)^2-2.\left(\frac{-1}{2}\right)^2+3.\left(\frac{-1}{2}\right)+1\)

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

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

\(=0\)

Bài 2: 

a) \(\frac{37-x}{x+13}=\frac{3}{7}\)

\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)

\(\Rightarrow259-7x=3x+39\)

\(\Rightarrow259-39=3x+7x\)

\(\Rightarrow220=10x\)

\(\Rightarrow x=22\)

d) \(\frac{3^2.3^8}{27^3}=3^x\)

\(\Rightarrow\frac{3^{10}}{\left(3^3\right)^3}=3^x\)

\(\frac{\Rightarrow3^{10}}{3^9}=3^x\)

\(\Rightarrow3=3^x\)

\(\Rightarrow x=1\)

Hok tốt nha^^

Bài 2:

a) \(\frac{37-x}{x+13}=\frac{3}{7}\)

\(\left(37-x\right).7=3\left(x+13\right)\)

\(259-7x=3x+39\)

\(259=3x+39+7x\)

\(259=10x+39\)

\(259-39=10x\)

\(220=10x\)

\(x=22\)

b) \(\left|x-\frac{1}{2}\right|-\sqrt{\frac{1}{9}}=\sqrt{\frac{1}{4}}\)

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

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

\(\left|x-\frac{1}{2}\right|=\frac{5}{6}\)

\(\orbr{\begin{cases}x-\frac{1}{2}=\frac{5}{6}\\x-\frac{1}{2}=-\frac{5}{6}\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{1}{3}\end{cases}}\)

c) \(2^{3x+2}=4^{x+5}\)

\(2^{3x+2}=2^{2\left(x+5\right)}\)

\(3x+2=2\left(x+5\right)\)

\(3x+2=2x+10\)

\(3x=2x+10-2\)

\(3x=2x+8\)

\(3x-2x=8\)

\(x=8\)

a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Leftrightarrow2x=18\)

\(\Leftrightarrow x=9\)

b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)

\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)

\(\Leftrightarrow x=66\)

b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)

\(\left(\frac{1}{3}\right)^{2.x}=\left(\frac{1}{3}\right)^{3.22}\)

\(2x=3.22\)

\(x=33\)

Có ai trả lời không zvậy ?!? -_-

\(\frac{2}{x+6}+\frac{1}{x+1}=\frac{3x+8}{x^2+7x+6}\left(ĐKXĐ:x\ne-6;-1\right)\)

\(\Leftrightarrow\frac{2\left(x+1\right)+x+6}{x^2+7x+6}=\frac{3x+8}{x^2+7x+6}\)

 \(\Leftrightarrow2x+2+x+6=3x+8\)

\(\Leftrightarrow3x+8=3x+8\)

          \(\Rightarrow\)x vô số nghiệm

Đk: x \(\ne\)0; x \(\ne\)\(\pm\)3

Ta có: A = \(\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)

A = \(\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{x^2+3\left(3-x\right)}{3\left(x+3\right)\left(3-x\right)}\)

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

A = \(\frac{-\left(x+3\right)}{x}\)

Để A < -1 <=> \(-\frac{\left(x+3\right)}{x}< -1\) <=> \(\frac{-x-3}{x}+1< 0\)

<=> \(\frac{-x-3+x}{x}< 0\) <=> \(-\frac{3}{x}< 0\) 

Do -3 <0 => x> 0

Vậy Để A < -1 <=> x > 0 và x khác 3