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a) 1/5.8+1/8.11+1/11.14+......+1/x.(x+3)=101/1540
1/3.3.[1/5.8+1/8,11+1/11.14+......+1/x.(x+3)=101/1540
1/3.[3/5.8+3/8.11+3/11.14+........+3/x.(x+3)]=101/1540
1/3.[1/5-1/8+1/8-1/11+1/11-1/14+....+1/x-1/x+3=101/1540
1/3.[1/5-1/x+3]=101/1540
1/5-1/x+3=101/1540.3
1/5-1/x+3=303/1540
1/x+3=1/3-303/1540=1/308
=>x+3=308 =>x=305
Vậy x=305
1/3.3(1/5.8+1/8.11+1/11.14+.....1/x(x+1)_101/1540
1/3.(1/5-1/8+1/8-1/11+1/11-1/14+....1/x+1/x+3)=101/1540
1/3.(1/5-1/x+3)=101/1540
1/5-1/x+3=101/1540/1/3=303/1540
1/x+3=1/5-303/1540=1/308
x+3+308
x=305
a)\(\left(-x-\dfrac{1}{9}\right)^2=\dfrac{4}{9}\)
\(\Rightarrow\left(-x-\dfrac{1}{9}\right)^2=\left(\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^2\)
*)Xét \(\left(-x-\dfrac{1}{9}\right)^2=\left(\dfrac{2}{3}\right)^2\)
\(\Rightarrow-x-\dfrac{1}{9}=\dfrac{2}{3}\Rightarrow-x=\dfrac{7}{9}\Rightarrow x=-\dfrac{7}{9}\)
*)Xét \(\left(-x-\dfrac{1}{9}\right)^2=\left(-\dfrac{2}{3}\right)^2\)
\(\Rightarrow-x-\dfrac{1}{9}=-\dfrac{2}{3}\Rightarrow-x=-\dfrac{5}{9}\Rightarrow x=\dfrac{5}{9}\)
b)\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=1\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)
\(\Rightarrow2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1991}{3986}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1993}\)
\(\Rightarrow x+1=1993\Rightarrow x=1992\)
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 \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
a) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=2\)
Vậy x=2
b) \(x+4=2^0+1^{2019}\)
\(\Leftrightarrow x+4=1+1\)
\(\Leftrightarrow x+4=2\)
\(\Leftrightarrow x=-2\)
Vậy x=-2