\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)

\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)

\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)

\(\frac{3}{2}x=\frac{10}{9}\)

\(x=\frac{10}{9}:\frac{3}{2}\)

\(x=\frac{20}{27}\)

Vậy x=\(\frac{20}{27}\)

\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)

\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)

\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)

\(\frac{9}{11}-x=\frac{-2}{11}\)

\(x=\frac{9}{11}-\frac{-2}{11}\)

\(x=1\)

Vậy x=1

\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)

\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)

\(\frac{-11}{12}\cdot x=\frac{21}{12}\)

\(x=\frac{-21}{11}\)

Vậy x=\(\frac{-21}{11}\)

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

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

\(\frac{3}{2}+x=\frac{23}{4}\)

\(x=\frac{17}{4}\)

Vậy x=\(\frac{17}{4}\)

\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)

\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)

\(\frac{3}{4}-x:\frac{2}{15}=-13\)

\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)

\(x:\frac{2}{15}=\frac{45}{4}\)

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

Vậy x=\(\frac{3}{2}\)

\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)

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

\(3-\left(\frac{1}{6}-x\right)=1\)

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

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

\(x=\frac{-11}{6}\)

Vậy x=\(\frac{-11}{6}\)

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

\(1-2x=\frac{-1}{10}\)

\(2x=1-\frac{-1}{10}\)

\(2x=\frac{11}{10}\)

\(x=\frac{11}{20}\)

Vậy x=\(\frac{11}{20}\)

\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)

\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\)                                                         \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)

\(\frac{1}{2}x=\frac{11}{12}\)                                                                        \(\frac{1}{2}x=\frac{-1}{4}\)

\(x=\frac{11}{6}\)                                                                              \(x=\frac{-1}{2}\)

Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)

\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)

\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)

\(\frac{3}{2}x=\frac{10}{9}\)

\(x=\frac{10}{9}:\frac{3}{2}\)

\(x=\frac{20}{27}\)

tk mình đi mình làm nốt cho hjhj ^^

nó lừa đấy bạn đừng nghe

Bài giải:

a, \(11.xx-66=4.x+11\)

\(11x^2-66=4.x+11\)

\(11x^2-66-4.x-11=0\)

\(11x^2-77-4x=0\)

\(11x^2-4x-77=0\)

\(x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4.11.\left(-77\right)}}{2.11}\)

\(x=\frac{4+\sqrt{16}+3388}{22}\)

\(x=\frac{4+\sqrt{3404}}{22}\)

\(x=\frac{4+2\sqrt{851}}{22}\)

\(x=\frac{2-\sqrt{851}}{11}\)

\(\Rightarrow\)Có hai trường hợp: \(x_1=\frac{2-\sqrt{851}}{11};x_2=\frac{2+\sqrt{851}}{11}\)

Tớ bận rồi, cậu coi câu trên đã nhé ! Tớ xin lỗi, khi nào tớ sẽ làm tiếp =)) 

dấu trừ đầu tiên các bạn thay thành số 4 hộ mik nhé

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

#)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) \(\left(x-11\right)+\frac{3x}{x-11}=3+\frac{33}{x-11}\)

\(\Leftrightarrow x+\frac{3x}{x-11}-\frac{33}{x-11}=14\)

\(\Leftrightarrow x^2-11x+3x-33=14x-154\)

\(\Leftrightarrow x^2-22x+121=0\)

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

Vậy .......

b) \(\frac{7-2x}{x-1}=\frac{1-4x}{x+2}\Leftrightarrow\left(7-2x\right)\left(x+2\right)=\left(1-4x\right)\left(x-1\right)\)

\(\Leftrightarrow7x-2x^2+14-4x=x-4x^2-1+4x\)

\(\Leftrightarrow2x^2=-15\)(vô lí)

Vậy pt vô nghiệm

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

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

\(\Leftrightarrow3x-2x^2-6x+4x=2x^2+2x-4x-4+x-4x^2+1-4x\)

\(\Leftrightarrow6x=-3\Leftrightarrow x=-\frac{1}{2}\)

Vậy.........

(gửi trước 3 câu)

d) \(\frac{109x-4}{111x+1}-1=0\Leftrightarrow109x-4=111x+1\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)

Vậy x=-5/2

e) \(\frac{x^2-7}{x}=x-\frac{1}{2}\Leftrightarrow\frac{x^2-7}{x}-\frac{x^2}{x}=-\frac{1}{2}\Leftrightarrow-\frac{7}{x}=\frac{1}{2}\Leftrightarrow x=-14\)

f) \(\frac{x+1}{x+2}=3\Leftrightarrow x+1=3x+6\Leftrightarrow2x=7\Leftrightarrow x=\frac{7}{2}\)

d) \(\frac{109x-4}{111x+1}=1\)

\(\Rightarrow\frac{109x-4}{111x+1}=1\)

\(\Rightarrow109x-4=111x+1\)

\(\Rightarrow-2x=5\)

\(\Rightarrow x=-2,5\)

Vậy x = -2,5

e) \(\frac{x^2-7}{x}=x-\frac{1}{2}\)

\(\Rightarrow\frac{x^2-7}{x}=\frac{2x-1}{2}\)

\(\Rightarrow\left(x^2-7\right)2=x\left(2x-1\right)\)

\(\Rightarrow2x^2-14=2x^2-x\)

\(\Rightarrow x=14\)

Vậy x = 14

f) \(\frac{x+1}{x+2}=3\)

\(\Rightarrow x+1=3\left(x+2\right)\)

\(\Rightarrow x+1=3x+6\)

\(\Rightarrow2x=-5\)

\(\Rightarrow x=-2,5\)

Vậy x = -2,5

\(x=\frac{903}{391}\)

Bài này sử dụng MTCT đó bạn!

903/391 mình nghĩ vậy

x = 903/391

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

Bài 3: 

 \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}=\dfrac{2}{3}+\dfrac{1}{3}=1\)