Bài 2: 

b: x+25%x=-1,25

=>1,25x=-1,25

hay x=-1

c: x-75%x=1/4

=>1/4x=1/4

hay x=1

Bài 2: 

a: =3/2-11/4=6/4-11/4=-5/4

b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3

\(\frac{\text{A=1,11+0,19-13.2}}{2,06+0,54}\) -(\(\frac{1}{2}\)

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, 3/4 + 1/4.x=2

    1/4.x        = 2-3/4

     1/4.x      =5/4

     x           = 5/4:1/4

     x           = 5

b, x-2/3.9/4=2,5-1/2

    x-2/3.9/4=2

    x-2/3     =2:9/4

    x-2/3   =8/9

     x        = 8/9+2/3

     x         = 14/9

c, 2.x+1/3=4/3

    2.x      =4/3-1/3

    2.x      =1

       x      =11:2

       x      = 1/2

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)

= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)

= \(x^8.\frac{1}{10}.1.1.1.1\)

= \(x^8.\frac{1}{10}\)

Mk ko pik co dung ko nua

=4/10

\(\frac{x^8}{10}\)

a)

\(\begin{array}{l}P + \frac{1}{{x + 2}} = \frac{x}{{{x^2} - 2{\rm{x}} + 4}}\\P = \frac{x}{{{x^2} - 2{\rm{x}} + 4}} - \frac{1}{{x + 2}}\\P = \frac{{x\left( {x + 2} \right) - {x^2} + 2{\rm{x}} - 4}}{{\left( {{x^2} - 2{\rm{x}} + 4} \right)\left( {x + 2} \right)}}\\P = \frac{{{x^2} + 2{\rm{x}} - {x^2} + 2{\rm{x}} + 4}}{{{x^3} + 8}}\\P = \frac{{4{\rm{x}} - 4}}{{{x^3} + 8}}\end{array}\)

b)

\(\begin{array}{l}P - \frac{{4\left( {x - 2} \right)}}{{x + 2}} = \frac{{16}}{{x - 2}}\\P = \frac{{16}}{{x - 2}} + \frac{{4\left( {x - 2} \right)}}{{x + 2}}\\P = \frac{{16\left( {x + 2} \right) + 4\left( {x - 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\P = \frac{{16{\rm{x}} + 32 + 4{{\rm{x}}^2} - 16{\rm{x}} + 16}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\P = \frac{{4{{\rm{x}}^2} + 48}}{{{x^2} - 4}}\end{array}\)

c) 

\(\begin{array}{l}P.\frac{{x - 2}}{{x + 3}} = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}\\ \Rightarrow P = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}.\frac{{x + 3}}{{x - 2}}\\P = \frac{{{{(x - 2)}^2}(x + 3)}}{{(x - 3)(x + 3)(x - 2)}} = \frac{{x - 2}}{{x - 3}}\end{array}\)\(\)

d)

\(\begin{array}{l}P:\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}} = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}\\ \Rightarrow P = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}.\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}}\\P = \frac{{(x - 2)(x + 2)(x - 3)(x + 3)}}{{2{\rm{x}}(x + 3)(x + 2)}}\\P = \frac{{(x - 2)(x - 3)}}{{2{\rm{x}}}}\end{array}\)

a) P=\(\dfrac{4x-4}{x^3-8}\)( lấy VP-VT)
b)P=\(\dfrac{4x^2+48}{x^2-4}\) ( chuyển VT và thành VP+VT)
c) P=\(\dfrac{x-2}{x-3}\) ( chuyển VT thành VP.VT là ra)
d) \(\dfrac{\left(x-2\right)\left(x-3\right)}{2x}\)( lấy VP.VT)

A= \(\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)

=\(-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)

B= \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)

b) ta có : A=\(-\frac{359}{390}\approx-0,9\)

B= \(\frac{13}{12}\approx1,08\)

=> A<x<B mà x nguyên => x=0 hoặc x=1

\(A=\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)

\(=-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)

\(B=\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)

Ta có : \(A=-\frac{359}{390}\approx-0,9\)

\(B=\frac{13}{12}\approx1,08\)

\(\Rightarrow A< x< B\) mà x nguyên \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Ta có:

\(A=\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{\frac{-13}{10}}{\frac{13}{5}}-\frac{5}{6}:2=\frac{-1}{2}-\frac{5}{12}=\frac{-11}{12}\)

\(B=\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}:\frac{75}{26}=\frac{13}{12}\)

\(\Rightarrow A< x< B\Rightarrow\frac{-11}{12}< x< \frac{13}{12}\Rightarrow-1< x\le1\Rightarrow x\in\left\{0;1\right\}\)