a) Tìm x biết: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
b) So sánh \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) với 1
c) Tìm GTNN của: A= |x-10|+|x-5|
K
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AH
Akai Haruma
Giáo viên
24 tháng 7 2021
Bạn tham khảo lời giải tại đây:
https://olm.vn/hoi-dap/detail/81621153379.html
10 tháng 10 2023
a) \(3\cdot24^{10}=3\cdot6^{10}\cdot4^{10}=3\cdot3^{10}\cdot2^{10}\cdot2^{20}\)
\(=3^{11}\cdot2^{30}\)
\(4^{30}=2^{30}\cdot2^{30}=2^{30}\cdot4^{15}\)
Ta có \(4^{15}>3^{15}>3^{11}\) nên \(4^{15}>3^{11}\)
Khi đó \(4^{15}\cdot2^{30}>3^{11}\cdot2^{30}\) hay \(4^{30}>3\cdot24^{10}\)
b) \(\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{19}{9^2\cdot10^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+...+\dfrac{19}{81\cdot100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
Vậy dãy trên nhỏ hơn 1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
10 tháng 10 2023
a/
\(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}.2^{30}=\left(2^2\right)^{15}.2^{30}=4^{15}.2^{30}\)
\(3.24^{10}=3.3^{10}.\left(2^3\right)^{10}=3^{11}.2^{30}< 3^{15}.2^{30}\)
\(\Rightarrow4^{30}=4^{15}.2^{30}>3^{15}.2^{30}>3^{11}.2^{30}=3.24^{10}\)
b/
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}=\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}=\)
\(=1-\dfrac{1}{10^2}< 1\)
D
20 tháng 1 2022
M=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{17}{8^2.9^2}+\dfrac{19}{9^2.10^2}\)
=\(\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{17}{64.81}+\dfrac{19}{81.100}\)
=\(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{81}+\dfrac{1}{81}-\dfrac{1}{100}\)
=1-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)<\(\dfrac{100}{100}=1\)
2 tháng 7 2021
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
30 tháng 9 2021
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\)
LL
2
15 tháng 8 2017
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{100^2}\right)\)
\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
15 tháng 8 2017
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
a/ \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
Mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
\(\Leftrightarrow x+101=0\)
\(\Leftrightarrow x=-101\)
Vậy...
b/ Đặt :
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+.........+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+....+\dfrac{10^2-9^2}{9^2.10^2}\)
\(=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+....+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{10^2}< 1\)
\(\Leftrightarrow A< 1\left(đpcm\right)\)
Vậy...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x-10\right|+\left|x-5\right|=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A\ge\left|x-10+5-x\right|\)
\(\Leftrightarrow A\ge5\)
Dấu "=" xảy ra
\(\Leftrightarrow\left(x-10\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge10\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le10\\5\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\5\le x\le10\end{matrix}\right.\)
Vậy..
đcmcm