tìm x biết :
x / 2018 + (x+2)/ 1010 + x - 2021 = 0
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Tìm x biết: x/2018 + x+2/1010 + x - 2021 = 0
Giúp mình đi mình tick cho ai đúng nhất và nhanh nhất.
Lưu ý: Giải chi tiết.
\(\frac{x}{2018}+\frac{x}{1010}+x-2021=0\)
\(\Rightarrow x\left(\frac{1}{2018}+\frac{1}{1010}-2021\right)=0\)
\(\Rightarrow x=0\)
P/s : ko chắc :v
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tìm x biết: x+1/2019+x+2/2018+x+3/2017=x-1/2021+x-2/2022+x-3/2023
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
Vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
=> x + 2020 = 0
=> x = -2020
Bài làm :
Ta có :
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
\(\text{Vì : }\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy x=-2020
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\frac{x+1+2019}{2019}+\frac{x+2+2018}{2018}+\frac{x+3+2017}{2017}=\frac{x-1+2021}{2021}+\frac{x-2+2022}{2022}+\frac{x-3+2023}{2023}\)\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Tính \(\left(x+y\right)^2-2021\). Biết x-y-1=0, x-2xy-y=-2018
Lời giải:
Ta có:
\(x-y-1=0\Rightarrow x-y=1\)
\(x-2xy-y=-2018\)
\(\Rightarrow 2xy=x-y+2018=1+2018=2019\)
Do đó:
\((x+y)^2-2021=x^2+2xy+y^2-2021\)
\(=x^2-2xy+y^2+4xy-2021\)
\(=(x-y)^2+4xy-2021=1^2+4.2019-2021=6056\)
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30 tháng 12 2017 lúc 21:22
1 tìm x biết
x-4/2021+x-3/2020=x-2/2019+x-1/2018
giúp mk vs lm ơn
Xem thêm câu trả lời
tìm x biết:
(x+1)/2018+(x+1)/2019+(x+1)/2020+(x+1)/2021
Giúp mk nhanh vs nhé các bn
Tìm x:
x-1/2020+x-2/2021=x+1/2018+x+2/2017
x−42021+x−32020=x−22019+x−12018" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
x−42021+x−32020−x−22019−x−12018=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">⇔ x+20172021+x+20172020−x+20172019−x+20172018=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
12021+12020−12019−12018)=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
⇔ x + 2017 = 0
⇔ x = -2017
\(\frac{x-1}{2020}+\frac{x-2}{2021}=\frac{x+1}{2018}+\frac{x+2}{2017}\)
\(\Leftrightarrow\frac{x-1}{2020}+1+\frac{x-2}{2021}-1=\frac{x+1}{2018}+1+\frac{x+2}{2017}+1\)
\(\Leftrightarrow\frac{x+2019}{2020}+\frac{x+2019}{2021}=\frac{x+2019}{2018}+\frac{x+2019}{2017}\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
mà \(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
Tìm giá trị lớn nhất của N=
\(\dfrac{-2\left|x-2018\right|-2021}{2020+\left|x-2018\right|}\)
Tìm x biết: x-4/2021 +x-3/2020= x-2/2019+x-1/2018
\(\dfrac{x-4}{2021}+\dfrac{x-3}{2020}=\dfrac{x-2}{2019}+\dfrac{x-1}{2018}\)
⇔ \(\dfrac{x-4}{2021}+\dfrac{x-3}{2020}-\dfrac{x-2}{2019}-\dfrac{x-1}{2018}=0\)
⇔ \(\left(1+\dfrac{x-4}{2021}\right)+\left(1+\dfrac{x-3}{2020}\right)-\left(1+\dfrac{x-2}{2019}\right)-\left(1+\dfrac{x-1}{2018}\right)=0\)⇔ \(\dfrac{x+2017}{2021}+\dfrac{x+2017}{2020}-\dfrac{x+2017}{2019}-\dfrac{x+2017}{2018}=0\)
⇔ \(\left(x+2017\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\right)=0\)
⇔ x + 2017 = 0
⇔ x = -2017
Vậy x = -2017
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Cho x,y là số thực thỏa mãn \(x^2+y^2+xy-3x-3y+3=0\). Chứng minh biểu thức P = \(\left(3x+2y-6\right)^{1010}+\left(x-y+1^{1011}\right)+2021\) có giá trị là một số nguyên