Tìm x
Ix+1/1.2I+Ix+1/2.3I+......+Ix+1/99.100I=100x
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LT
1
16 tháng 7 2018
\(|x+\frac{1}{1\cdot5}|+|x+\frac{1}{5\cdot9}|+|x+\frac{1}{9\cdot13}|+...+|x+\frac{1}{379\cdot401}|=101x\)
Ta có:
\(|x+\frac{1}{1\cdot5}|\ge0\forall x\)
\(|x+\frac{1}{5\cdot9}|\ge0\forall x\)
\(|x+\frac{1}{9\cdot13}|\ge0\forall x\)
\(......\)
\(|x+\frac{1}{397\cdot401}|\ge0\forall x\)
\(\Rightarrow|x+\frac{1}{1\cdot5}|+|x+\frac{1}{5\cdot9}|+|x+\frac{1}{9\cdot13}|+...+|x+\frac{1}{397\cdot401}|\ge0\)
\(\Rightarrow\left(x+\frac{1}{1\cdot5}\right)+\left(x+\frac{1}{5\cdot9}\right)+\left(x+\frac{1}{9\cdot13}\right)+...+\left(x+\frac{1}{397\cdot401}\right)=101x\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{397\cdot401}\right)=101x\)
\(\Rightarrow100x+\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{397\cdot401}\right)=101x\)
Đặt \(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{397\cdot401}\)
\(\Rightarrow4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{397\cdot401}\right)\)
\(\Rightarrow4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{397\cdot401}\)
\(\Rightarrow4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{397}-\frac{1}{401}\)
\(\Rightarrow4A=1-\frac{1}{401}\)
\(\Rightarrow4A=\frac{400}{401}\)
\(\Rightarrow A=\frac{400}{401}:4\)
\(\Rightarrow A=\frac{400}{401}\cdot\frac{1}{4}\)
\(\Rightarrow A=\frac{100}{401}\)
\(\Rightarrow100x+\frac{100}{401}=101x\)
\(\Rightarrow101x-100x=\frac{100}{401}\)
\(\Rightarrow x=\frac{100}{401}\)
Vậy \(x=\frac{100}{401}\)
18 tháng 9 2017
Ta có:
1.Ix+1I + Ix+2I + Ix+3I + ... Ix+12I=11x
=> x>=0
=>x+1 + x+2 + x+3 + ... x+12=11x
=> (x+x+x+x..+x)+(1+2+...+12)=11x
Dãy 1;2;...;12 có số số hạng là:
(12-1)+1=12 ( số hạng )
=> (12x)+(12+1).12:2=12x+78=11x
=> -x=78
=> x=-78
k bít có đúng k
ND
0
\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|=100x\)
=> \(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)
=> \(\left[x+x+x+...+x\right]+\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{99\cdot100}\right]=100x\)
=> \(99x+\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right]=100x\)
=> \(99x+\left[1-\frac{1}{100}\right]=100x\)
=> \(99x+\frac{99}{100}=100x\)
=> \(100x-99x=\frac{99}{100}\)
=> \(x=\frac{99}{100}\)
Check lại có đúng không nhé