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

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

             \(.\)                   \(.\)

             \(.\)

             \(.\)                    \(.\)  

             \(.\)                    \(.\)

         \(\frac{1}{2013^2}< \frac{1}{2012\cdot2013}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{2013^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}\)

Mà \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}=1-\frac{1}{2013}< 1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{2013^2}< 1\)

Nhớ k cho mình nhé!

Chúc các bạn học tốt!

mình giải ở đè trước rồi

A) \(\frac{10}{12}\)+\(2\)- /\(\frac{-2}{3}\)/ -\(\frac{3}{4}\)= \(\frac{10}{12}\)+2-\(\frac{2}{3}\)-\(\frac{3}{4}\)= \(\frac{10}{12}\)+\(\frac{24}{12}\)-\(\frac{8}{12}\)-\(\frac{9}{12}\)=\(\frac{17}{12}\)

tương tự bài B= \(\frac{59}{40}\)

mk hk bk ghi dáu GTTĐ nên mk ghi như thế 

bạn tính kết quả trong dấu GT tuyệt đối rồi bạn mở dấu GTTĐ bằng cách cho số đó trở thành số dương là được

chúc bn may mắn

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

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

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

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

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

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

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

Vậy \(x=\frac{5}{4}\)

bai 1

\(\frac{7}{4}\)+ \(\frac{5}{6}\):5 - 0,375.2.\(^{\left(-2\right)^2}\)= \(\frac{7}{4}\)+ \(\frac{5}{6}\)x\(\frac{1}{5}\)- \(\frac{15}{4}\). 2.4=\(\frac{7}{4}\)+\(\frac{1}{6}\)-\(\frac{15}{4}\).8=\(\frac{42}{24}\)+\(\frac{4}{24}\)-30=\(\frac{11}{6}\)-30=-169/6

\(\frac{1}{4}\)+\(\frac{3}{4}\). \(\left(\frac{-1}{2}+\frac{2}{3}\right)\)=\(\frac{1}{4}\)+ \(\frac{3}{4}\).\(\left(\frac{-3}{6}+\frac{4}{6}\right)\)= \(\frac{1}{4}+\frac{3}{4}.\frac{1}{6}=\frac{1}{4}+\frac{3}{8}\)= \(\frac{5}{8}\)

1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200

 

A=1/2+1/3+1/6+1/12
A=(1/2+1/3+1/6)+1/12
A=1+1/12
A=13/12

B=20/3+61/8+10/3+37/8
B=30/3+98/8
B=10+49/4
B=89/4

      a) \(\frac{x}{3}\)\(-\)\(\frac{1}{8}\)\(=\)\(\frac{5}{8}\)

\(\Rightarrow\) \(\frac{x}{3}\)=\(\frac{5}{8}+\frac{1}{8}\)

\(\Rightarrow\)  \(\frac{x}{3}\)\(=\frac{3}{4}\)

\(\Rightarrow\)  x=3.3:4

\(\Rightarrow\) x=2,25

Vậy x=2,25

     b) \(3\frac{1}{3}\).x-\(6\frac{3}{4}=3\frac{1}{4}\)

\(\Rightarrow\)\(\frac{10}{3}\).x-\(\frac{27}{4}=\frac{13}{4}\)

\(\Rightarrow\) \(\frac{10}{3}\).x=\(\frac{13}{4}+\frac{27}{4}\)

\(\Rightarrow\) \(\frac{10}{3}\).x=10

\(\Rightarrow\) x=\(10\div\frac{10}{3}\)

\(\Rightarrow\) x=3

Vậy x=3

Ủng hộ mik nha mọi người !

Ta có : \(\frac{x}{3}-\frac{1}{8}=\frac{5}{8}\)

\(\Leftrightarrow\frac{x}{3}=\frac{5}{8}-\frac{1}{8}\)

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

\(\Leftrightarrow x=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

\(\frac{x}{3}-\frac{1}{8}+\frac{5}{8}\)

\(\frac{x}{3}=\frac{5}{8}+\frac{1}{8}=\frac{6}{8}=\frac{3}{4}\)

\(x=\frac{3}{4}.3=\frac{9}{4}\)

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

\(\frac{10}{3}x=3\frac{1}{4}+6\frac{3}{4}=10\)

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

\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)

\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)

\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)

Ko hiểu đề!

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

<=>\(\left(\frac{7}{2}+2x\right).\frac{8}{3}=\frac{16}{3}\)

<=>\(\frac{28}{3}+\frac{16x}{3}=\frac{16}{3}\)

<=>\(\frac{16x}{3}=\frac{-2}{3}\)

<=>\(16x=-2\)

<=>\(x=\frac{-1}{8}\)

vậy \(x=\frac{-1}{8}\)

b,\(\left|2x+3\right|=5\)

xét x<0,ta co: \(\left|2x+3\right|=5\)<=> \(-2x+3=5\)<=>\(-2x=2\)<=>\(x=-1\)(loại)

xét x>0,ta co:\(\left|2x+3\right|=5\)<=>\(2x+3=5\)<=>\(2x=2\)<=>\(x=1\)

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

<=>\(\frac{3x-6}{12}=\frac{20+4x}{12}\)

=>\(3x-6=20+4x\)

<=>\(3x-6-20-4x=0\)

<=>\(-x-26=0\)

<=>\(-x=26\)

<=>\(x=-26\)

kl:.......