Question

1. Tìm số nguyên \(|x|\) = 7

2. Tính nhanh giá trị biểu thức M = \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

Answer 1

1. Ta có: \(\left|x\right|=7\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

Vậy \(x\in\left\{\pm7\right\}\)

2. \(M=\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)

\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{13}{39}-\dfrac{3}{39}\right)\)

\(\Rightarrow M=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{1.10}{2.39}=\dfrac{5}{39}\)

Tick mk vs! Thank nhiều!

Answer 2

1. Theo đb ta có: |x|=7
=> Có 2 TH:\(\left\{{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\) \(\in Z\)
Vậy x=7 \(\veebar\) x= -7 ( x\(\in\) Z) thì |x|=7
2. \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Đặt A= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Ta thấy: \(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
\(\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{2}{5.7}\)
... \(\dfrac{1}{11}-\dfrac{1}{13}=\dfrac{2}{11.13}\)
=> 2D=2(\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\))
<=> 2D= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
<=>2D=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
<=> 2D= \(\dfrac{1}{3}-\dfrac{1}{13}\)
<=>2D= \(\dfrac{13}{39}-\dfrac{3}{39}\)
<=>2D=\(\dfrac{10}{39}\)
=> D= \(\dfrac{10}{39}:2\)
<=> D= \(\dfrac{10}{39}.\dfrac{1}{2}\)
<=> D=\(\dfrac{5}{39}\)
Vậy D= \(\dfrac{5}{39}\)
_ Chc bn hk tốt_