a, x( x - 3) - ( x + 2)( x - 1) = 3

<=> x² - 3x - x² + x - 2x + 2 = 3

<=>  x² - 3x - x² + x - 2x = 3 - 2

<=> -4x = 1

<=> x =\(-\frac{1}{4}\)

Vậy_

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

\(\Leftrightarrow\frac{30x}{30}-\frac{\left(x-1\right)6}{30}+\frac{\left(x+2\right)5}{30}=\frac{120}{30}+\frac{\left(x+1\right)10}{30}\)

=> 30x - 6x + 6 + 5x + 10 = 120 + 10x + 10

<=> 30x - 6x + 5x - 10x = 120 + 10 - 6 - 10

<=> 19x = 114

<=> x = 6

Vậy _

Bài 1: 

\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\) 

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

                                                  Vậy B=5

Bài 2:

a) x³ - 36x = 0  

=>  x(x²-36)=0

=>  x(x²+6x-6x-36)=0 

=> x[x(x+6)-6(x+6) ]=0

=> x(x+6)(x-6)=0

\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)

 \(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)

                                  Vậy x=0; x=-6; x=6

b)  (x - y = 4 => x=4+y)

 x−3y−2 =32  

=>2(x-3) = 3(y-2)

=>2x-6= 3y-6

=>2x-3y=0

=>2(4+y)-3y=0

=>8+2y-3y=0

=>8-y=0

=>y=8 (thỏa mãn)

Do đó x=4+y=4+8=12 (thỏa mãn)

         Vậy x=12 và y =8

B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4  1/5 - 1/8 

B= 1/ 1/2 + 3

B= 2+3

B=5

B2:

a) x^3 - 36x = 0

x(x^2 - 36) = 0

=> x=0  hoặc x^2-36=0

=> x= 0 hoặc x^2=36

=> x=0 hoặc x= +- 6

b) x-y = 4 => x= 4+y

thay x=4+y vào x- 3/ y-2=3/2, có:

4+y-3/ y+2 = 3/2

y+1/ y+2 = 3/2

y+2 -1/ y+2 = 3/2

1 - 1/y+2 = 3/2

1/y+2= 1-3/2

1/y+2 = -1/2

=> y+2 = -2

=> y= -4

Dp x= 4+y => x= 4-4

=> x=0

Vậy x=0 và y=-4

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

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

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

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

a) \(\frac{a}{b}x-\frac{7}{8}=\frac{1}{4}\)

\(\Rightarrow\frac{a}{b}x=\frac{1}{4}+\frac{7}{8}\)

\(\Rightarrow\frac{a}{b}x=\frac{9}{8}\)

\(\Rightarrow x=\frac{9}{8}:\frac{a}{b}=\frac{9}{8}.\frac{b}{a}\)

\(\Rightarrow x=\frac{9b}{8a}\)

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

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

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

\(\Rightarrow\frac{3}{2}x=\frac{1}{10}\)

\(\Rightarrow x=\frac{1}{10}:\frac{3}{2}\)

\(\Rightarrow x=\frac{1}{15}\)

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

\(\Rightarrow\frac{2}{3}x+\frac{5}{6}-\frac{2}{9}+\frac{1}{3}x=\frac{4}{3}\)

\(\Rightarrow\frac{2}{3}x+\frac{1}{3}x=\frac{4}{3}-\frac{5}{6}+\frac{2}{9}\)

\(\Rightarrow x=\frac{13}{18}\)

a) \(\frac{3}{4}+\frac{1}{4}.x=\frac{1}{2}+\frac{1}{2}x\)

\(\Rightarrow3.\frac{1}{4}+\frac{1}{4}.x=\frac{1}{2}.\left(x+1\right)\)

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

\(\Rightarrow\frac{x+1}{x+3}=\frac{1}{4}:\frac{1}{2}=\frac{1}{2}\)\(\Rightarrow\left(x+1\right).2=x+3\Rightarrow2x+2=x+3\)

\(\Rightarrow2x-x=3-2\Rightarrow x=1\)

vay x=1

Cho mình sửa lại đề câu 1b: \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\frac{x}{7}-\frac{1}{2}=\frac{1}{y+1}\)

\(\frac{2x-7}{14}=\frac{1}{y+1}\)

\(TH1:\hept{\begin{cases}2x-7=7\\y+1=2\end{cases}\Rightarrow\hept{\begin{cases}x=7\\y=1\end{cases}}}\)

\(TH2:\hept{\begin{cases}2x-7=-7\\y+1=-2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=-3\end{cases}}}\)

nhớ cho