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

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 => \(B< \frac{7}{8}< 1\)

\(\Rightarrow B< 1\left(đpcm\right)\)

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

có chỗ x^7 hả

a) \(x=\frac{9}{10}\)

b) \(x=\frac{-4}{3}\)

c) \(x=\frac{1}{42}\)

d) \(x=\frac{-47}{10}\)

ko có thời gian nên mình chỉ cho đáp án thôi nhé

thông cảm cho mình ngen

đúng thì k đấy

chúc bạn học giỏi

làm chi tiết cho mk nhé

ai làm chi tiết mk k cho nhìu

\(a,\frac{3}{7}+\left|2x-\frac{1}{2}\right|=\frac{4}{5}\)

\(\Rightarrow\left|2x-\frac{1}{2}\right|=\frac{13}{35}\)

\(\Rightarrow2x-\frac{1}{2}=\pm\left(\frac{13}{35}\right)\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{13}{35}\\2x-\frac{1}{2}=\frac{-13}{35}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{61}{70}\\2x=\frac{9}{70}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{61}{140}\\x=\frac{9}{140}\end{cases}}\)

~Study well~

#KSJ

\(b,\frac{3}{4}-4\times\left|2x+1\right|=\frac{1}{2}\)

\(\Rightarrow4\times\left|2x+1\right|=\frac{1}{4}\)

\(\Rightarrow\left|2x+1\right|=\frac{1}{16}\)

\(\Rightarrow2x+1=\pm\left(\frac{1}{16}\right)\)

\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{1}{16}\\2x+1=\frac{-1}{16}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{-15}{16}\\2x=\frac{-17}{16}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{32}\\x=\frac{-17}{32}\end{cases}}\)

~Study well~

#KSJ

\(c,\left(\frac{1}{2}+\frac{3}{4}\right)\times\left|3x-1\right|=\frac{1}{2}\)

\(\Rightarrow\frac{5}{4}\times\left|3x-1\right|=\frac{1}{2}\)

\(\Rightarrow\left|3x-1\right|=\frac{2}{5}\)

\(\Rightarrow3x-1=\pm\left(\frac{2}{5}\right)\)

\(\Rightarrow\orbr{\begin{cases}3x-1=\frac{2}{5}\\3x-1=\frac{-2}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=\frac{7}{5}\\3x=\frac{3}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{15}\\x=\frac{1}{5}\end{cases}}\)

~Study well~

#KSJ

\(\frac{ }{ }\)

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

\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)

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

\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)

\(\Leftrightarrow x=\frac{-2}{5}\)

\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)

\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)

Đến đây tìm được rồi nhé

3,4, áp dụng bài 1,2 rồi làm :v

1. 1/2x - 3/5 = -4/5

=> 1/2x = -4/5 + 3/5

=> 1/2x = -1/5

=> x = -1/5 : 1/2

=> x = -2/5