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B=(1+1/1 x 3) x (1+1/2 x 4) x (1+1/3 x 5) x..x (1+1/2020 x 2022)
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NT
1
Những câu hỏi liên quan
ND
Nguyễn Đức Trí
VIP
8 tháng 7 2023
\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)
= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)
= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)
= \(\dfrac{337.2023}{2}\)
= \(\dfrac{\text{681751}}{2}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
13 tháng 12 2023
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 11 2023
c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0
(\(x\) - 2022).(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)
1 tháng 10 2021
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
HK
16 tháng 4 2023
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
NH
22 tháng 2 2023
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
10 tháng 7 2019
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)
Số nào xuất hiện 2 lần thì thay thế những số đó bằng số 1.
\(B=\frac{1}{2020}\)
XO
10 tháng 7 2019
B = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right).\left(1-\frac{1}{2020}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2018}{2019}.\frac{2019}{2020}\)
= \(\frac{1.2.3...2019}{2.3.4..2020}\)(Nếu có 2 thừa số giống nhau lặp lại ở tử số và mẫu số thì rút gọn coi như triệt tiêu hết và không có gì)
= \(\frac{1}{2020}\)
NH
5 tháng 10 2023
\(1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\left(đk:x>2\right)\)
đặt
\(A=1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\)
\(\left(x-1\right)^2A=\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\)
\(\left(x-1\right)^2A-A=\left[\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\right]-\left[1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\right]\)
\(\left[\left(x-1\right)^2-1\right]A=\left(x-1\right)^{2022}-1\)
\(A=\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}\)
\(=>\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\\ =>\left(x-1\right)^{2022}-1=17^{2022}-1\\ =>\left(x-1\right)^{2022}=17^{2022}\\ =>x-1=17\\ =>x=18\left(tm\right)\)
B
1
\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2020\cdot2022}\right)\)
\(=\left(1+\dfrac{1}{2^2-1}\right)\cdot\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2021^2-1}\right)\)
\(=\dfrac{2^2-1+2}{2^2-1}\cdot\dfrac{3^2-1+1}{3^2-1}\cdot...\cdot\dfrac{2021^2-1+1}{2021^2-1}\)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2021^2}{\left(2021-1\right)\left(2021+1\right)}\)
\(=\dfrac{2\cdot3\cdot...\cdot2021}{1\cdot2\cdot3\cdot...\cdot2020}\cdot\dfrac{2\cdot3\cdot...\cdot2021}{3\cdot4\cdot...\cdot2022}\)
\(=\dfrac{2021}{1}\cdot\dfrac{2}{2022}=\dfrac{2021}{1011}\)