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a) Đặt \(A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{\left(x-2\right)x}+\frac{1}{x\left(x+2\right)}\)
=> \(3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{\left(x-2\right)x}+\frac{3}{x\left(x+2\right)}\)
=> \(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-2\right)}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+2}\)
=> 3A = \(\frac{1}{5}-\frac{1}{x+2}\)
=> A = \(\frac{1}{15}-\frac{1}{3x+6}\)
Mà : A = \(\frac{101}{1540}\)
=> \(\frac{1}{15}-\frac{1}{3x+6}=\frac{101}{1540}\)
=> \(\frac{1}{3x+6}=\frac{1}{15}-\frac{101}{1540}=\frac{1}{924}\)
=> 3x + 6 = 924
=> 3(x + 2) = 924
=> x + 2 = 308
=> x = 306
a) Ta có: \({{1} \over x(x+2)}= {{1} \over 3}({{1} \over x}-{{1} \over x+2})\) \(\Rightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over 8}+{{1} \over 8}-...+{{1} \over x}-{{1} \over x+2})={{101} \over 1540} \)\(\Leftrightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over x+2})={{101} \over 1540}\)\(\Leftrightarrow\)x+2 = 308 \(\Leftrightarrow\) x=306 Lúc sau lm hơi tắt mọi người thông cảm
Tìm x:
1+\(\frac{1}{3}\)+\(\frac{1}{6}\)+....+\(\frac{2}{x.\left(x+1\right)}\)=\(1\frac{1991}{1993}\)
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=1+1-\frac{2}{x+1}=2-\frac{2}{x+1}\)
Do đó ta có phương trình:
\(2-\frac{2}{x+1}=1\frac{1991}{1993}\)
<=> \(\frac{2}{1993}=\frac{2}{x+1}\)
<=> x + 1 = 1993
<=> x = 1992
a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 3
x = 305
b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.\frac{3984}{1993}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)
=> x + 1 = 1993
x = 1993 - 1
x = 1992
a ,\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
Vậy x = 305
a, \(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+...+\(\dfrac{1}{x\left(x+3\right)}\)=\(\dfrac{101}{1540}\)
\(\dfrac{1}{3}\)(\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)+\(\dfrac{3}{11.14}\)+...+\(\dfrac{3}{x\left(x+3\right)}\))=\(\dfrac{101}{1540}\)
\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)
\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)
\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\)=\(\dfrac{101}{1540}\) : \(\dfrac{1}{3}\)
\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\)=\(\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}\)=\(\dfrac{1}{5}\)-\(\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}\)=\(\dfrac{1}{308}\)
<=>1(x+3)=308.1
<=>1(x+3)=308
<=> x+3=308:1
<=> x+3=308
<=> x=308-3
<=> x=305
b,1+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{1}{x\left(x+1\right):2}\)=1\(\dfrac{1991}{1993}\)
\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{3984}{1993}\)\(2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3984}{1993}\)
\(2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)
\(2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)
\(1-\dfrac{1}{x+1}=\dfrac{3984}{1993}:2\)
\(1-\dfrac{1}{x+1}=\dfrac{1992}{1993}\)
\(\dfrac{1}{x+1}=1-\dfrac{1992}{1993}\)
\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)
<=>1(x+1)=1993.1
<=>1(x+1)=1993
<=> x+1=1993 : 1
<=> x+1=1993
<=> x=1993-1
<=> x=1992
ko có kết quả biểu thức này ra bao nhiêu thì sao lm đc
a)Ta có \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
\(VT=2(1-\frac{1}{x+1})\). Do đó : \(2(1-\frac{1}{x+1})=1\frac{1989}{1991}\)
\(2(\frac{x+1-1}{x+1})=1\frac{1989}{1991}\)
\(\frac{x}{x+1}=\frac{3980}{1991}:2\)
\(\frac{x}{x+1}=\frac{1990}{1991}\)
Vậy x = 1990
\(VT=2\left(1-\frac{1}{x+1}\right)\) . Do đó : \(2\left(1-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(2\left(\frac{x+1-1}{x+1}\right)=1\frac{1989}{1991}\)
\(\frac{x}{x+1}=\frac{3980}{1991}\)
\(\frac{x}{x+1}=\frac{1990}{1991}\)
Vậy x = 1990