Tìm x biết : X x (1+1/3+1/6+1/10+1/15+...+1/190)=2013x19/10
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tìm x biết \(|x+1|+|x+\frac{1}{3}|+|x+\frac{1}{6}|+|x+\frac{1}{10}|+...+|x+\frac{1}{190}|=20x\) =20x
Ta có \(\left|x+1\right|\ge0;\left|x+\frac{1}{3}\right|\ge0;...;\)\(\left|x+\frac{1}{190}\right|\ge0\) \(\forall x\)
=> \(\left|x+1\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{190}\right|\ge0\) \(\forall x\)
=> \(20x\ge0\Rightarrow x\ge0\)
Với \(x\ge0\) => \(x+1>0,x+\frac{1}{3}>0,x+\frac{1}{6}>0,...,x+\frac{1}{190}>0\)
=> \(\left|x+1\right|=x+1,\left|x+\frac{1}{3}\right|=x+\frac{1}{3},\left|x+\frac{1}{6}\right|=x+\frac{1}{6},...,\left|x+\frac{1}{190}\right|=x+\frac{1}{190}\)
=> \(x+1+x+\frac{1}{3}+x+\frac{1}{6}+...+x+\frac{1}{190}=20x\)
=> \(19x+\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)=20x\)
=> \(x=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)\)
Gọi \(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{190}\)
=> \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)
=> \(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
=> \(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
=> \(\frac{1}{2}A=1-\frac{1}{20}\)
=> \(A=\frac{19}{10}\)
Thay vào ta có
=> \(x=-\frac{19}{10}\)
dạng chuỗi nha bạn
ko hiểu thì tích cho mình là mình giải cho
\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right).a=1\)
=> \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}.a=1\)
=> \(\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}.a=1\)
=> \(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}.a=1\)
=> \(\frac{1.2.3.4...38}{2.3.4.5...39}.\frac{4.5.6.7...41}{3.4.5.6...40}.a=1\)
=> \(\frac{1}{39}.\frac{41}{3}.a=1\)
=> \(\frac{41}{117}.a=1\)
=> \(a=1:\frac{41}{117}=\frac{117}{41}\)
Bui Bao Ngoc oi! phan a,la (x+1)/10 hay x+(1/10)?
cac phan khac nua!
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)
\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)
Em giải như XYZ olm em nhé
Sau đó em thêm vào lập luận sau:
\(x\) = \(\dfrac{11}{18}\)
Vì \(\in\) N*
Vậy \(x\in\) \(\varnothing\)