1+³⁺⁴⁺⁹⁼

Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2016}{2017}\)

\(=\frac{1.2.3......2016}{2.3.4.......2017}\)

\(=\frac{1}{2017}\)

g)=>x+1/2=0

x=0-1/2

x=-1/2

hoặc 2/3-2x=0

2x=2/3-0

2x=2/3

x=2/3:2

x=1/3

nhìn @_@ hoa cả mắt đăng từng bài thôi bạn

Tìm x :

a) \(2.x.\frac{-3}{4}=-\frac{5}{12}\)

\(\Rightarrow2x=-\frac{5}{12}:-\frac{3}{4}\)

\(\Rightarrow2x=\frac{5}{9}\)

\(\Rightarrow x=\frac{5}{9}:2\)

\(\Rightarrow x=\frac{5}{18}\)

Vậy : \(x=\frac{5}{18}\)

b) \(\frac{2}{3}+\frac{1}{3}.x=7\)

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

\(\Rightarrow\frac{1}{3}.x=\frac{19}{3}\)

\(\Rightarrow x=\frac{19}{3}:\frac{1}{3}\)

\(\Rightarrow x=19\)

Vậy : \(x=19\)

c) \(\left(4.x+\frac{1}{8}\right)=\frac{3}{10}\)

\(\Rightarrow4.x=\frac{3}{10}-\frac{1}{8}\)

\(\Rightarrow4.x=\frac{7}{40}\)

\(\Rightarrow x=\frac{7}{40}:4\)

\(\Rightarrow x=\frac{7}{160}\)

Vậy : \(x=\frac{7}{160}\)

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

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

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

\(\Rightarrow\frac{1}{3}.x=\frac{13}{2}\)

\(\Rightarrow x=\frac{13}{2}:\frac{1}{3}\)

\(\Rightarrow x=\frac{39}{2}\)

Vậy : \(x=\frac{39}{2}\)

e) \(-\frac{2}{3}.x+\frac{1}{3}=-\frac{1}{2}\)

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

\(\Rightarrow-\frac{2}{3}.x=-\frac{5}{6}\)

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

\(\Rightarrow x=\frac{5}{4}\)

Vậy : \(x=\frac{5}{4}\)

0,7.2\(\frac23\).20.0,375.\(\frac{5}{28}\)

= \(\frac{7}{10}\) .\(\frac83\).20.\(\frac38\).\(\frac{5}{28}\)

= (\(\frac{7}{10}\).\(\frac{5}{28}\).20).(\(\frac83\).\(\frac38\))

= (\(\frac18.20\)). 1

= \(\frac52\)

B = (9,75 .21\(\frac37\) + \(\frac{39}{4}\).18\(\frac47\))

B = (\(\frac{39}{4}\).\(\frac{150}{7}\frac{}{7}\)+ \(\frac{39}{4}\).\(\frac{130}{7}\))

B = \(\frac{39}{4}\).(\(\frac{150}{7}\) + \(\frac{130}{7}\))

B = \(\frac{39}{4}\).\(\frac{280}{7}\)

B = 390

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

...

\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)

Cộng vế theo vế 

\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)

\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)

Lại có \(\frac{7}{8}< 1\)

Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)

\(\Rightarrow B< 1\left(đpcm\right)\)

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

a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)

hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)

Vậy: \(x=\frac{9}{10}\)

b) Ta có: \(5\frac{4}{7}:x=13\)

\(\Leftrightarrow\frac{39}{7}:x=13\)

\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)

Vậy: \(x=\frac{3}{7}\)

c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=84\)

\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)

Vậy: x=30

d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

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

hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)

Vậy: x=-5

e) Ta có: \(8\frac{2}{3}:x-10=-8\)

\(\Leftrightarrow\frac{26}{3}:x=2\)

hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)

Vậy: \(x=\frac{13}{3}\)

g) Ta có: \(x+30\%=-1.3\)

\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)

hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)

Vậy: \(x=\frac{-8}{5}\)

i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)

\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)

\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)

\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)

Vậy: x=-9

k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)

hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)

Vậy: x=30

m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)

\(\Leftrightarrow\left|2x-1\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)

thank you nha!

a) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)

b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)

c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)

d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)

e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)

Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((

a) \(\frac{9}{20}\)                                                   c) \(\frac{-55}{4}\)                                                                                                                                 

b) \(\frac{116}{75}\)                                                d) \(\frac{-76}{45}\)

đúng hết đấy nhé mình tính kĩ lắm ko sai đâu

       chúc may mắn

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}