tìm x biết :
1/1.2.3 + 1/2.3.4 + 1/3.4.5 + …. + 1/x(x+1)(x+2) = 1998/1999
K
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NT
0
NT
0
DK
0
MH
1
30 tháng 3 2017
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\right)x=-3\)
\(\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)x=-3\)
\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)x=-3\)
\(\dfrac{1}{2}.\dfrac{4949}{9900}x=-3\)
\(\dfrac{4949}{19800}x=-3\)
\(x=-3:\dfrac{4949}{19800}\)
\(x=-\dfrac{59400}{4949}\)
PN
1
16 tháng 9 2015
Ở sbt 6 tập mấy ko nhớ có bài tương tự trong ngoặc, mở phần lời giải ra để tính trong ngoặc nha
NV
0
12 tháng 1 2017
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
5 tháng 4 2017
đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
\(A=\frac{22}{45}:2=\frac{11}{45}\)
thay A vào ta được
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
XO
13 tháng 8 2019
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{1998}{1999}\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{1998}{1999}\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{1998}{1999}\)
\(\Leftrightarrow\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1998}{1999}\div\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{3996}{1999}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{1.2}-\frac{3996}{1999}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{-5993}{3998}\)
Như kiểu đề sai hay sao í