lỗi

x + 4/9 + 8/18 + 12/27 + 16/36 = 16/9

 x + (4/9+12/27)+(8/12+16/36) = 16/9

 x +       8/9         +     10/9        = 16/9

  x +               2                         = 16/9

  x                                              = 16/9 - 2

  x                                               =  -2/18  

x+4/9+4/9+4/9+4/9=16/9
x+4/9.4=16/9
x+16/9=16/9
-->    x = 0

a: =-21/36-3/36=-24/36=-2/3

b: =43/12*1/2+5/24=43/24+5/24=2

c: =8/9+1/9=1

e: =1-1/4+1/4-1/7+...+1/97-1/100

=1-1/100=99/100

a. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{-5}{48}\)

\(\dfrac{-1}{12}< x< \dfrac{1}{8}\) hay \(-0,08333...< x< 0,125\)

Vì \(x\in Z\Rightarrow x\in\left\{0\right\}\)

Ta có:

\(\dfrac{1}{4}+\dfrac{8}{9}=\dfrac{41}{36}\)

\(1-\left(\dfrac{3}{8}-\dfrac{5}{6}\right)=\dfrac{35}{24}\)

ta lại có:

\(\dfrac{41}{36}=\dfrac{82}{72}\)

\(\dfrac{35}{24}=\dfrac{105}{72}\)

\(\Rightarrow\)82 \(\le\)2x < 105

\(\Rightarrow\)41\(\le\)x<52.5

Vậy 41\(\le\)x<52.5

mọi người giúp mik với, mai nộp rùi

a)\(\dfrac{7}{-25}+-\dfrac{8}{25}=-\dfrac{15}{25}=-\dfrac{3}{5}\)
b)\(\dfrac{7}{21}-\dfrac{9}{-36}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)
c)\(-\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{1}{4}+\dfrac{5}{7}\)
\(=\left(-\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
\(=-\dfrac{1}{2}+1\)
\(=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}\)

\(a,\dfrac{7}{-25}+\dfrac{-8}{25}=\dfrac{-7}{25}+\dfrac{-8}{25}=\dfrac{-15}{25}=\dfrac{-3}{5}\\ b,\dfrac{7}{21}-\dfrac{9}{-36}=\dfrac{7}{21}+\dfrac{9}{36}=\dfrac{7}{12}\\ c,\dfrac{-3}{4}+\dfrac{2}{7}+\dfrac{1}{4}+\dfrac{5}{7}\\ =\left(\dfrac{-3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\\ =-\dfrac{1}{2}+1\\ =\dfrac{1}{2}\)

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

\(\dfrac{3}{8}\left(x-\dfrac{2}{3}\right)-\dfrac{2}{9}\left(3x-\dfrac{1}{2}\right)=-\dfrac{7}{36}\)

\(\Leftrightarrow\dfrac{3}{8}x-\dfrac{1}{4}-\dfrac{2}{3}x+\dfrac{1}{9}=-\dfrac{7}{36}\)

\(\Leftrightarrow\dfrac{-7}{24}x=-\dfrac{1}{18}\)\(\Leftrightarrow x=\dfrac{4}{21}\)

Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.

b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)

=>\(\dfrac{-3}{5}.x=1\)

=>\(x=1:\dfrac{-3}{5}\)

=>\(x=\dfrac{-5}{3}\)

Vậy \(x=\dfrac{-5}{3}\)

Cách 1

\(2\dfrac{3}{4}\)+ \(5\dfrac{7}{9}\)= \(\dfrac{11}{4}+\dfrac{47}{9}\)

= \(\dfrac{99}{36}+\dfrac{188}{36}\)

= \(\dfrac{287}{36}\)= 7,9722...

Cách 2

\(2\dfrac{3}{4}+5\dfrac{7}{8}=2\dfrac{55}{36}+5\dfrac{55}{36}\)

= \(7\dfrac{55}{36}=8\dfrac{19}{36}\)

Cách 1 mình làm nhầm bài đúng phải là

\(2\dfrac{3}{4}+5\dfrac{7}{9}=\dfrac{11}{4}+\dfrac{52}{9}=\dfrac{99}{36}+\dfrac{208}{36}\)

=\(\dfrac{307}{36}\)=8,527777778