Bài 2 : 

a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)

b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)

a, x + 7/12 = 17/18 - 1/9

    x + 7/12 = 15/18

                x = 15/18 - 7/12

                  x = 1/4

b, 29/30 - ( 13/23 + x ) = 7/69

( 13/23 + x ) = 29/30 - 7/69

13/23 + x = 199/230

              x = 199/230 - 13/23

                x = 3/10

hok tốt

x= 1372221/18559

(x-99) (1/30 + 1/32 + 1/34 - 1/36 - 1/38) = 0

SUy ra x - 99 = 0

VẬy x =99

\(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}\)

\(\Leftrightarrow\frac{x-69}{30}+\frac{x-67}{62}+\frac{x-65}{34}-\frac{x-63}{36}-\frac{x-61}{38}=0\)

\(\Leftrightarrow\frac{x-69-30}{30}+\frac{x-67-32}{32}+\frac{x-65-34}{34}-\frac{x-63-36}{36}+\frac{x-61-38}{38}=0\)

\(\Leftrightarrow\frac{x-99}{30}+\frac{x-99}{32}+\frac{x-99}{34}-\frac{x-99}{36}-\frac{x-99}{38}=0\)

\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}\right)=0\)

Vì \(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}\ne0\)

\(\Rightarrow x-99=0\)

\(\Rightarrow x=99\)

chịu

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{3}{2}\le x\le\dfrac{37}{24}-\dfrac{3-16}{24}=\dfrac{37-3+16}{24}=\dfrac{50}{24}=\dfrac{25}{12}\)

=>3/2<=x<=25/12

mà x là số nguyên

nên x=2

b: \(\Leftrightarrow-\dfrac{1}{23}-\dfrac{3}{23}-\dfrac{7}{23}< x\le\dfrac{1}{23}-\dfrac{8}{23}\)

=>-11<x<=-7

mà x là số nguyên

nên \(x\in\left\{-10;-9;-8;-7\right\}\)

a: =>19/23>19/x>19/29

=>\(x\in\left\{24;25;26;27;28\right\}\)

b: =>88/132<88/x<88/128

=>132>x>128

=>\(x\in\left\{131;130;129\right\}\)

c: =>\(\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{x}{8}< 0\\\dfrac{x}{8}-\dfrac{5}{x}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{32-x^2}{8x}< 0\\\dfrac{x^2-40}{8x}< 0\end{matrix}\right.\)

=>32<x^2<40

=>x=6

d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7

\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>x^2+11x+28=54

=>x^2+11x-26=0

=>(x+13)(x-2)=0

=>x=2 hoặc x=-13

e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)

=>x-258=0

=>x=258

1: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)

=5/13+3

=5/13+39/13

=44/13

2: \(B=\dfrac{2^{15}\cdot3^{30}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{3^3\cdot2\cdot2^{14}\cdot3^{14}\cdot3^{14}-2^2\cdot3\cdot2^{15}\cdot7^5}\)

\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)

\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\cdot\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)