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\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
\(\begin{array}{l} a)\left( {x + 1} \right) + \left( {x + 3} \right) + \left( {x + 5} \right) + ... + \left( {x + 99} \right) = 0\\ \Leftrightarrow 50x + \left( {1 + 3 + 5 + ... + 99} \right) = 0\\ \Leftrightarrow 50x + \left( {99 + 1} \right).25 = 0\\ \Leftrightarrow 50x + 2500 = 0\\ \Leftrightarrow x = - 50 \end{array}\)
\(\begin{array}{l} b)\left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) + ... + 10 + 11 = 11\\ \Leftrightarrow \left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) + \left( {1 + 2 + 3 + ... + 10} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) + 55 = 0\\ \Leftrightarrow \left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) = - 55\\ \Leftrightarrow 3x = - 49\\ \Leftrightarrow x = - \dfrac{{49}}{3} \end{array}\)
B=5+2(x-2019)2020
Vì (x-2019)2020 ≥0
=>5+(x-2019)2020 ≥5
Để B đạt Min
=>x-2019=0
=>x=2019
Vậy MinB=5 <=>x=2019
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
3M=3+3^2+...+3^2019
=>2M=3^2019-1
=>M=1/2(3^2019-1)
\(x+3^{2019}=\dfrac{4}{2}\left(3^{2019}-1\right)+5\)
=>x+3^2019=2*3^2019-2+5=2*3^2019+3
=>x=3^2019+3
a) (x+3)(x+5)=0
=>x+3=0 hoặc x+5=0
=>x=-3 hoặc -5
b) (x-1).5-1=0
=>5x-5-1=0
=>5x-6=0
=>5x=6
=>x=6/5
c)
làm câu c,d,b và E đi bạn