Toán Lớp 7: Tìm x: |x+1/2|+|x+1/6|+|x+1/12|+|x+1/20|+…+|x+1/110|=11x
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Toán Lớp 7: Tìm x: |x+1/2|+|x+1/6|+|x+1/12|+|x+1/20|+…+|x+1/110|=11x
Comments ( 2 )
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+…\left|x+\frac{1}{110}\right|=11x\)
\(\Leftrightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+…\left|x+\frac{1}{110}\right|\ge0\)
\(\rightarrow11x\ge0\rightarrow x\ge0\)
\(\text{Ta có:}\)
\(x+\frac{1}{2}+…+x+\frac{1}{110}=11x\)
\(\rightarrow10x+\frac{10}{11}=11x\)
\(\rightarrow x=\frac{10}{11}\)
|x+1/2|+|x+1/6|+|x+1/12|+|x+1/20|+…+|x+1/110| = 11x (1)
\forall x ta có :
|x+1/2| \ge0
|x+1/6| \ge0
|x+1/12| \ge0
|x+1/20| \ge 0
…..
|x+1/110| \ge 0
=> |x+1/2|+|x+1/6|+|x+1/12|+|x+1/20|+…+|x+1/110| \ge 0
Mà |x+1/2|+|x+1/6|+|x+1/12|+|x+1/20|+…+|x+1/110| = 11x nên 11x \ge 0
=>x \ge 0
Với x \ge 0 thì (1) trở thành :
(x + 1/2) + (x+1/6) + (x+1/12) + (x+1/20) + … + (x+1/110) = 11x
=> x + 1/2 + x + 1/6 + x + 1/12 + x + 1/20 + ….. + x + 1/110 = 11x
=> \underbrace{(x + x + x + ….+x)} + (1/2 +1/6 + 1/12 + 1/20 + … + 1/110) = 11x
\text{có 10 số hạng}
=> 10x + (1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + …. + 1/(10.11) ) = 11x
=> 10x + (1 – 1/2 + 1/2- 1/3 + 1/3 – 1/4 + 1/4 -1/5 + … + 1/10 – 1/11) = 11x
=> 10x + (1 – 1/11) = 11x
=> 10x + 10/11 = 11x
=> x = 10/11
Vậy x=10/11 là giá trị cần tìm.