3B=3/1.4+3/4.7+3/7.10+...+3/100.103

3B=(4-1)/1.4+(7-4)/4.7+(10-7)/7.10+...+(103-100)/100.103

3B=1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103=1-1/103=102/103

B=102/(3.103)=34/103

HT

\(\frac{1}{1.4}+\frac{1}{4.7}+.....+\frac{1}{100.103}\)

Đặt :

\(A=\frac{1}{1.4}+\frac{1}{4.7}+....+\frac{1}{100.103}\)

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{100.103}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{100}-\frac{1}{103}\)

\(3A=1-\frac{1}{103}\)

\(3A=\frac{102}{103}\)

\(A=\frac{34}{103}\)

giúp mình với ToT

trả lời nhớ k nhé!

=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1999\cdot2000}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...-\frac{1}{999}+\frac{1}{999}-\frac{1}{2000}\)

=\(\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{999}+\frac{1}{999}\right)-\frac{1}{2000}\)

=\(\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)

=\(\frac{1}{1}-\frac{1}{2000}\)

=\(\frac{2000}{2000}-\frac{1}{2000}\)

=\(\frac{1999}{2000}\)

Ta có :

b - a = 57

\(\frac{a}{-9}=\frac{b}{10}\)

Áp dụng tính chất dãy tỉ số bănhg nhau , ta có :

\(\frac{a}{-9}=\frac{b}{10}=\frac{b-a}{10-\left(-9\right)}=\frac{57}{19}=3\)

\(\Rightarrow\hept{\begin{cases}a=3.\left(-9\right)=-27\\b=3.10=30\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
a/-9=b/10=b-a/10+9=57/19=3
=>a/-9=3=>a=-27
=>b/10=3=>b=30

   *CHÚC BẠN HỌC TỐT*

Ta có :

\(\frac{-11}{6}< x< \frac{20}{13}\)

\(\Rightarrow-1\frac{5}{6}< x< 1\frac{7}{13}\)

\(\Rightarrow-2< x< 2\)

=> x ∈ { -1 ; 0 ; 1 } ( do x ∈ Z )

Ta có: \(\frac{x}{3}=\frac{y}{4}=k\)

\(\Rightarrow\)\(x=3k\)\(;\)\(y=4k\)

Ta lại có: \(x.y=12\)hay   \(3k.4k=12\)

\(\Rightarrow\)\(12k^2=12\Rightarrow k^2=1\Rightarrow k\pm1\)

Với \(k=1\), ta có \(x=3\)\(;\)\(y=4\)

Với \(k=-1\), ta có \(x=-3\)\(;\)\(y=-4\)

Trả lời:

\(\frac{-5}{8}-x=\frac{3}{7}\)

\(x=\frac{-5}{8}-\frac{3}{7}\)

\(x=-\frac{59}{56}\)