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co \(\frac{1}{9\cdot10}=\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{10\cdot11}=\frac{1}{10}-\frac{1}{11}\)
............
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
nen \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-...+\frac{1}{x}-\frac{1}{x+1}\)
=\(\frac{1}{9}-\frac{1}{x+1}\)
2 . ( \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\))
= 2 . ( \(\frac{1}{9}-\frac{1}{x+1}\)) = \(\frac{2}{9}-\frac{2}{x+1}\)
Ta có A=1/10.11+1/11.12+...+1/98.99+1/99.100
=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100
=1/10-1/100
=10/100-1/100
=9/100
Vậy A=9/100
Giải:
A=1/10.11+1/11.12+...+1/98.99+1/99.100
A=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100
A=1/10-1/100
A=9/100
Chúc bạn học tốt!
1) A=7/10.11+7/11.12+7/12.13+...+7/69.70
A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)
A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)
A= 7.(1/10-1/70)
A=7.3/35=3/5
2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75
B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75
B=1/25-1/75=2/75
A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70
(1/7).A=1/10.11+1/11.12+...+1/69.70
=1/10-1/11+1/11-1/12+...+1/69-1/70
=1/10-1/70=3/35
=>A=7.(3/35)
=3/5
2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75
=>(1/2).B=2/25.27+...+2.73.75
=1/25-1/27+...+1/73-1/75
=1/25-1/75
=2/75
=>B=4/75
\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\div2\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=18-1\)
\(\Leftrightarrow x=17\)
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Leftrightarrow\left|x\right|=\frac{20}{12}+\frac{9}{12}\)
\(\Leftrightarrow\left|x\right|=\frac{29}{12}\)
\(\Leftrightarrow x=\pm\frac{29}{12}\)
đặt a=1/10.11+ 1/11.12+..+1/99.100
=1/10-1/11+1/11-1/12+...+1/99-1/100
=1/10-1/100=9/100
vậy a=9/100
|x+3|=|-9|
TH1: x+3=9 => x=9-3 TH2: x+3=-9=> x=-9 -3
x=6 x=-12