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Tính nhanh.\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{57.40}\)
\(=5.\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{4}{37.40}\right)\)
\(=\frac{5}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{40}\right)\)
\(=\frac{5}{3}\left(\frac{1}{1}-\frac{1}{40}\right)\)
\(=\frac{5}{3}.\frac{39}{40}\)
\(=\frac{13}{8}\)
Rút gobj p/s
\(\frac{2019.2020+4038}{2022.2011-4044}\)
\(=\frac{2019.\left(2020+2\right)}{2020.\left(2011-2\right)}\)
\(=\frac{2019.2022}{2022.2019}\)
\(=\frac{1}{1}=1\)
Study well
Cho mk sorry nha dong thứ 2 từ trên cuống dưới phải là
\(5.\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{37.40}\right)\) nha
Sorry nhiều
Study well
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ta có :\(E=\frac{2019^{2019}+1}{2019^{2020}+1}\Leftrightarrow2019\cdot E=\frac{2019^{2020}+2019}{2019^{2020}+1}=1+\frac{2019}{2019^{2020}+1}\)
\(F=\frac{2019^{2020}+1}{2019^{2021}+1}\Leftrightarrow2019\cdot F=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)
vì \(\frac{2019}{2019^{2020}+1}>\frac{2019}{2019^{2021}+1}\) nên E>F
E=2019 x 2019 x 2019 x ........ x 2019 x2019 +1 /2019 x 2019 x 2019 x.........x 2019 x 2019 + 1
E=1+1/2019+1
E=2/2020
E=1/1010
F=2019 x 2019 x 2019 x .......... x 2019 x 2019 +1 / 2019 x 2019 x 2019 x ....... x 2019 x 2019 +1
F= 1+1/2019+1
F=2/2020
F=1/1010
từ đó ta có E=F(=1/1010)
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(x+2019)(x-2020)=0.
=> x+2019=0 hoặc x-2020=0.
+, x+2019=0. +, x-2020=0
x= 0-2019 x = 0+2020
x = -2019. x = 2020.
Vậy: x thuộc{ -2019 ; 2020 }.
#Học tốt.
\(\Leftrightarrow\orbr{\begin{cases}x+2019=0\\x-2020=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2019\\x=2020\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có công thức A.B=0 suy ra A=0,B=0
Suy ra X-2019=0 ⟹X=0+2019 ⟹X=2019
X-2020=0 ⟹X=0+2020 ⟹X=2020
\(\left(x-2019\right).\left(x-2020\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2019=0\\x-2020=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2019\\x=2020\end{cases}}\)
Vậy \(x=2019\)hoặc \(x=2020\)
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1. Tự làm
2. Ta có: \(x_1+x_2+x_3+...+x_{2017}+x_{2018}+x_{2019}+x_{2020}=0\)
=> \(\left(x_1+x_2+x_3\right)+\left(x_4+x_5+x_6\right)+....+\left(x_{2017}+x_{2018}+x_{2019}\right)+x_{2020}=0\)
=> \(3+3+....+3+x_{2020}=0\) (gồm 673 chữ số 3 vì x1 + .... + x2019 gồm 2019 hạng tử gộp lại mỗi cặp 3 hạng tử)
=> \(3.673+x_{2020}=0\)
=> \(2019+x_{2020}=0\)
=> \(x_{2020}=-2019\)
3. a) 3(x - 1) - (x - 5) = -18
=> 3x - 3 - x + 5 = -18
=> 2x + 2 = -18
=> 2x = -18 - 2
=> 2x = -20
=> x = -20 : 2
=> x = 10
b ) x + (x + 1) + (x + 2) + ... + (x + 2019) = 0
=> (x + x + ... + x) + (1 + 2 + ... + 2019) = 0
=> 2020x + (2019 + 1).[(2019 - 1) : 1 + 1] : 2 = 0
=> 2020x + 2020. 2019 : 2 = 0
=> 2020x + 2039190 = 0
=> 2020x = -2039190
=> x = -2039190 : 2020
=> x = -10095
(xem lại đề)
c) Ta có: 3x + 23 = 3(x + 4) + 11
Do 3(x + 4) \(⋮\)4 => 11 \(⋮\)x + 4
=> x + 4 \(\in\)Ư(11) = {1; -1; 11; -11}
Với: +) x + 4 = 1 => x = 1 - 4 = -3
+) x + 4 = -1 => x = -1 - 4 = -5
+) x + 4 = 11 => x = 11 - 4 = 7
+) x + 4 = -11 => x = -11 - 4 = -15
4a) Ta có: 22x - y = 21x + x - y = 21 + (x - y)
Do 21x \(⋮\)7; x - y \(⋮\)7
=> 22x - y \(⋮\)7
b) 8x + 20y = 7x + 21y + x - y = 7(x + 3y) + (x - y)
Do : 7(x + 3y) \(⋮\)7; x - y \(⋮\)7
=> 8x + 20y \(⋮\)7
c) 11x + 10y = 14x + 7y - 3x + 3y = 7(2x + y) - 3(x - y)
Do: 7(2x + y) \(⋮\)7; 3(x - y) \(⋮\)7
=> 11x + 10y \(⋮\)7
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(M=\frac{2019\times2020-2}{2018+2018\times2020}=\frac{2019\times2020-2}{2018+2018\times2020+2020-2020}=\frac{2019\times2020-2}{\left(2018+1\right)\times2020+2018-2020}=\frac{2019\times2020-2}{2019\times2020-2}=1\\ N=\frac{-2019\times20202020}{20192019\times2020}=\frac{-2019\times10001\times2020}{2019\times10001\times2020}=-1\)
b)\(5\left|x-1\right|=3M-2N=5\\ \left|x-1\right|=1\Rightarrow\hept{\begin{cases}x-1=1\Rightarrow x=2\\x-1=-1\Rightarrow x=0\end{cases}}\)
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237.(-26)+26.137
=(-237).26+26.137
=26(-237+137)
=26.100
=2600
63.(-25)+25.(-23)
=-63.25+25.(-23)
=25.(-63+(-23))
=25.86
=2150
-2.(-3).(-2014)<0
(-1).(-2)....(-2014)>0
ko hiểu chỗ nào nhắn cho mình
\(\left(x-2019\right)\left(x-2020\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2019=0\\x-2020=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2019\\x=2020\end{cases}}\)
Vậy \(x\in\left\{2019;2020\right\}\)