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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
Ta có: \(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
\(\Rightarrow P=\left(1+\frac{1}{49}\right)+\left(1+\frac{2}{48}\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+1\)
\(\Rightarrow P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(\Rightarrow P=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}}{50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)}=\frac{1}{50}\)
Vậy \(\frac{S}{P}=\frac{1}{50}\)
bài 1 :
\(\frac{2}{3}\)+\(\frac{1}{3}\)=\(\frac{3}{3}\)=1
\(\frac{3}{4}\)+\(\frac{2}{4}\)+\(\frac{1}{4}\)=\(\frac{4}{4}\)=1
\(\frac{4}{5}\)+\(\frac{3}{5}\)+\(\frac{2}{5}\)+\(\frac{1}{5}\)=\(\frac{10}{5}\)= 2
chúc bạn học tốt !!!
\(a,\frac{x+5}{20}=-\frac{5}{4}\)
\(\left(x+5\right):20=\frac{-5}{4}\)
\(x+5=-\frac{5}{4}\times20\)
\(x+5=-25\)
\(x=-25-5\)
\(x=-30\)
các câu khác thì theo đó tự làm nha
bn ơi các câu a, b thì mk làm đc nhưng vướng câu c đó bn ạ
a) \(\frac{5}{6}=\frac{x-1}{x}\)
\(5x=6x-6\)
\(6x-5x=6\)
\(x=6\)
các câu còn lại lm tương tự
hok tốt!!
b) \(\frac{1}{2}=\frac{x+1}{3x}\)
\(\Rightarrow1.3x=2.\left(x+1\right)\)
\(3x=2x+2\)
\(3x-2x=2\)
\(x=2\)
Vậy x=2
các câu khác bạn làm tương tự
a)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2-\left(\frac{3}{5}\right)^2=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}+\frac{3}{5}\right)\left(2x+\frac{3}{5}-\frac{3}{5}\right)=0\)
\(\Leftrightarrow\left(2x+\frac{6}{5}\right).2x=0\)
\(\Leftrightarrow\left[\begin{matrix}x=-\frac{3}{5}\\x=0\end{matrix}\right.\)
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b) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{19}=0\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{19}:3\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{57}\)
\(\Leftrightarrow3x-\frac{1}{2}=\sqrt[3]{-\frac{1}{57}}\)
\(\Leftrightarrow3x=\sqrt[3]{-\frac{1}{57}}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{\sqrt[3]{-\frac{1}{57}}+\frac{1}{2}}{3}\)
Số hơi to
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