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Câu 1:
c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)
\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)
\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)
Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)
=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)
=>x=10
Câu 2:
d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)
\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)
\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\Rightarrow\hept{\begin{cases}a=2020k\\b=2021k\\c=2022k\end{cases}}\)
Khi đó M = 4(a - b)(b - c) - (c - a)2
= 4(2020k - 2021k)(2021k - 2022k) - (2022k - 2020k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 4k2 = 0
Vậy M = 0
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\)( \(k\ne0\))
\(\Rightarrow a=2020k\); \(b=2021k\); \(c=2022k\)
Thay a, b, c vào biểu thức M ta có:
\(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(2020k-2021k\right)\left(2021k-2022k\right)-\left(2022k-2020k\right)^2\)
\(=4.\left(-k\right).\left(-k\right)-\left(2k\right)^2=4k^2-4k^2=0\)
Vậy \(M=0\)
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\)
\(=1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}\)
\(=3+Q\)
Suy ra \(3+Q=1\Leftrightarrow Q=-2\).
\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2020}\)
\(\Leftrightarrow(\frac{x+4}{2019}+1)+(\frac{x+3}{2020}+1)=(\frac{x+2}{2021}+1)+(\frac{x+1}{2022}+1)\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{x+2023}{2022}\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)
\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2023=0\)
\(\Leftrightarrow x=-2023\)
a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé
a) \(3^{x+1}=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
b) \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\left(\frac{1}{2}\right)^6\)
\(\Leftrightarrow x+1=6\Leftrightarrow x=5\)
c) \(\frac{81}{3x}=9\)
\(\Leftrightarrow3x=9\Leftrightarrow x=3\)
d) \(2^{x+1}+2^{x+2}=192\)
\(\Leftrightarrow2^x.2+2^x.4=192\)
\(\Leftrightarrow2^x.6=192\Leftrightarrow2^x=32\Leftrightarrow x=5\)
e) Ta có : \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}\Rightarrow\left(x-1\right)^{2020}+\left(y+2\right)^{2020}\ge0}\)
Mà \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Bài giải
a, \(3^{x+1}=243\)
\(3^{x+1}=3^5\)
\(\Rightarrow\text{ }x+1=5\)
\(\Rightarrow\text{ }x=4\)
b, \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\frac{1}{2^{x+1}}=\frac{1}{2^6}\)
\(2^{x+1}=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
c, \(\frac{81}{3x}=9\)
\(27x=81\)
\(x=3\)
d, \(2^{x+1}+2^{x+2}=192\)
\(2^{x+1}\left(1+2\right)=192\)
\(2^{x+1}\cdot3=192\)
\(2^{x+1}=64=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
e, \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
Mà \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}}\) với mọi x,y nên \(\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
\(\Rightarrow\text{ }x=1\text{ ; }y=-2\)
mình nghĩ là 2022
nếu đúng thì cho mình xin 1 tick
Ta có:
\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2021}\)
\(C=\left(\right.1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{2021}\left.\right)-\left(\right.\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2020}\left.\right)\)
Mà
\(1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{2021}=\left(\right.1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2021}\left.\right)-\left(\right.\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2020}\left.\right)\)
\(=\left(\right.1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2021}\left.\right)\)
(Vì tổng các số lẻ = tổng tất cả − tổng các số chẵn)
và
\(2\left(\right.\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2020}\left.\right)=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1010}\)
Suy ra
\(C=\left(\right.1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2021}\left.\right)-\left(\right.1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1010}\left.\right)\)
Do đó \(C=\frac{1}{1011}+\frac{1}{1012}+\ldots+\frac{1}{2021}\)
Mặt khác \(D=\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2022}\)\(\)\(\)\(\)\(\)\(\)\(\)
nên \(C - D = \frac{1}{1011} - \frac{1}{2022}\)
⇒ \(C - D = \frac{2}{2022} - \frac{1}{2022} = \frac{1}{2022}\)
Vậy \(I=\left\lbrack\left(\frac{1}{2022}\right)\right\rbrack^{2021}+2022\)
ta biến đổi D từ C
\(D=\left(1+\frac12+\frac13+\cdots+\frac{1}{2022}\right)-\left(1+\frac12+\frac13+\frac14+.\ldots+\frac{1}{2011}\right)\)
\(D=\left(1+\frac12+\frac13+\cdots+\frac{1}{2022}\right)-\left(\frac22+\frac24+\frac26+\frac28+.\ldots+\frac{2}{2022}\right)\)
\(D=1+\left(\frac12-\frac22\right)+\frac13+\left(\frac14-\frac24\right)+.\ldots+\left(\frac{1}{2022}-\frac{2}{2022}\right)\)
\(D=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2011}-\frac{1}{2022}\)
mik nghĩ đề bị lỗi rồi:v, phải bỏ \(\frac{1}{2022}\) của D ra chứ nếu làm tiếp là:
=> \(C-D=\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}\right)-\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2011}-\frac{1}{2022}\right)\)
\(C-D=\frac{1}{2022}\)
=> \(I=\left(\frac{1}{2022}\right)^{2021}+2022\)
bạn kiểm tra mik có vt thừa gì ko nhá chứ bên mik thấy cứ lạ lạ
C = 1 - 1/2 + 1/3 - 1/4 + ... + 1/2021
C = 1/1011 + 1/1012 + ... + 1/2021
D = 1/1012 + 1/1013 + ... + 1/2021 + 1/2022
C - D = 1/1011 - 1/2022 = 1/2022
I = (C - D)^2021 + 2022
I = (1/2022)^2021 + 2022
Vậy I = 2022 + 1/2022^2021, vì các phân số từ 1/1012 đến 1/2021 triệt tiêu nhau.