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nhân cả 2 vế của đẳng thức với 1/2 ta được
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)
\(\frac{1}{x+1}=-\frac{2013}{4030}\)
hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)
\(x+1=-\frac{4030}{2013}\)
\(=>x=-\frac{6043}{2013}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)
\(\Leftrightarrow x=216-1=215\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
\(3x-5x+x=\frac{1}{5}+3+\frac{3}{2}\)
\(-x=\frac{47}{10}\)
\(x=\frac{-47}{10}\)
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)
\(a,\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x-1=24\)
\(\Rightarrow x=25\)
\(b,-\frac{x}{4}=-\frac{9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
\(c,\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x^2+x=72\)
\(\Leftrightarrow x\left(x+1\right)=72..\)
ấn nhầm: lm tiếp nhé!
\(x\left(x+1\right)=72\)
\(\text{Mà x thuộc Z nên }x\left(x+1\right)=8\left(8+1\right)\)
\(\Leftrightarrow x=8\)
\(\Rightarrow\frac{4x^2-4x+1}{3}-\frac{3}{2}\left(x^2+6x+9\right)=\frac{1}{3}\left(x^2-1\right)+2x\)
\(\Rightarrow\frac{4x^2-4x+1}{3}-\frac{3x^2+18x+27}{2}=\frac{x^2-1}{3}+2x\)
\(\Rightarrow8x^2-8x+2-9x^2-54x-81=2x^2-2+12x\)
\(\Rightarrow-3x^2-74x-77=0\)
\(\Delta=5476-4.\left(-77\right).\left(-3\right)=4552\)
\(\Rightarrow\sqrt{\Delta}=\sqrt{4552}\)
\(\Rightarrow x=\frac{-74+\sqrt{4552}}{6};x=\frac{-74-\sqrt{4552}}{6}\)
\(\frac{\left(2x-1\right)^2}{3}-\frac{3.\left(x+3\right)^2}{2}=\frac{x^2-1}{3}+2x\)
Qui đồng lên là tìm được
a) \(\left|x+\frac{1}{2}\right|=\frac{1}{3}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)
Vậy....
b) \(\left|x-\frac{1}{2}\right|=\frac{1}{3}-\frac{1}{2}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\) vô lí do \(\left|a\right|\ge0\)
Vậy pt vô nghiệm
c) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\)\(\left|x+\frac{1}{3}\right|=3\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)
Vậy..
d) \(\left|x-\frac{1}{5}\right|+\frac{1}{3}=\frac{1}{4}-\left|-\frac{3}{2}\right|\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|+\frac{1}{3}=-\frac{5}{4}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do \(\left|a\right|\ge0\)với mọi a
Vậy pt vô nghiệm
e) \(\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left(\frac{2}{3}-\frac{1}{2}\right)\)
\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\frac{2}{3}\\x=\frac{4}{3}\end{cases}}\)
Vậy...
\(\Rightarrow4x-\left(\frac{4}{5.7.9}+\frac{4}{7.9.11}+...+\frac{4}{99.101.103}\right)=\frac{2}{83224}=\frac{1}{41612}\)
\(4x-\left(\frac{9-5}{5.7.9}+\frac{11-7}{7.9.11}+...+\frac{103-99}{99.101.103}\right)=\frac{1}{41612}\)
\(4x-\left(\frac{1}{5.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.11}+...+\frac{1}{99.101}-\frac{1}{101.103}\right)=\frac{1}{41612}\)
\(4x-\left(\frac{1}{5.7}-\frac{1}{101.103}\right)=\frac{1}{41612}\)
Từ đó tìm ra x