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Lời giải:
a.
PT $\Leftrightarrow (x+3)^2=2016^{2020}-17^{91}+9$
Ta thấy: $2016^{2020}-17^{91}+9\equiv 0-(-1)^{91}+0\equiv -1\equiv 2\pmod 3$
Mà 1 scp thì chia $3$ chỉ dư $0$ hoặc $1$ nên pt vô nghiệm.
b.
$x^2=2016(y-1)^2-2017^{2019}\equiv 0-1^{2019}\equiv 3\pmod 4$
Mà 1 scp chia $4$ chỉ dư $0$ hoặc $1$ nên vô lý.
Vậy pt vô nghiệm.
c.
$(x-1)^2=2017^{2017}+1\equiv 1^{2017}+1\equiv 2\pmod 4$
Mà 1 scp khi chia cho $4$ chỉ dư $0$ hoặc $1$ nên vô lý
Vậy pt vô nghiệm
d.
$(x+2)^2=2018^{10}+4\equiv (-1)^{10}+1\equiv 2\pmod 3$
Mà 1 scp khi chia $3$ dư $0$ hoặc $1$ nên vô lý
Vậy pt vô nghiệm.
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a, TK:
(x lẻ do \(2y^2-8y+3=2\left(y^2-4y\right)+3=x^2\) lẻ)
\(b,\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+4y+4\right)=9\\ \Leftrightarrow\left(x-2\right)^2+\left(y+2\right)^2=9\)
Vậy pt vô nghiệm do 9 ko phải tổng 2 số chính phương
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\(\dfrac{2-x}{2017}+1=\dfrac{x-1}{2018}-1+1-\dfrac{x}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{x-2019}{2018}+\dfrac{2019-x}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}+\dfrac{2019-x}{2018}-\dfrac{2019-x}{2019}=0\)
\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)
\(\Leftrightarrow2019-x=0\) (do \(\dfrac{1}{2017}>\dfrac{1}{2019}\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}>0\))
\(\Rightarrow x=2019\)
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\(\frac{2-x}{2017}-1=\frac{1-x}{2018}-\frac{x}{2019}\)
\(\Leftrightarrow\) \(\frac{2-x}{2017}+1=\frac{1-x}{2018}+1-\frac{x}{2019}+1\)
\(\Leftrightarrow\) \(\frac{2019-x}{2017}=\frac{2019-x}{2018}-\frac{2019-x}{2019}\)
\(\Leftrightarrow\) \(\frac{2019-x}{2017}-\frac{2019-x}{2018}+\frac{2019-x}{2019}=0\)
\(\Leftrightarrow\) \(\left(2019-x\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\right)=0\)
Mà \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\right)\ne0\)
\(\Rightarrow\) \(2019-x=0\) \(\Leftrightarrow\) \(x=2019\)
\(\Rightarrow\) \(S=\left\{2019\right\}\)
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\(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\Leftrightarrow\left(\dfrac{2-x}{2017}+1\right)=\left(\dfrac{1-x}{2018}+1\right)+\left(1-\dfrac{x}{2019}\right)\)
\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{2019-x}{2018}+\dfrac{2019-x}{2019}\)\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)
Ta đã có: \(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}< 0\)
Vậy ta dễ dàng suy ra được \(S=\left\{2019\right\}\)
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buithianhtho làm cách này mà ko có máy tính thì đến bao giờ ?
\(\dfrac{x-3}{2017}+\dfrac{x-2}{2018}+\dfrac{x-1}{2019}=3\)
\(\Leftrightarrow\dfrac{x-3}{2017}-1+\dfrac{x-2}{2018}-1+\dfrac{x-1}{2019}-1=3-1-1-1\)
\(\Leftrightarrow\dfrac{x-3-2017}{2017}+\dfrac{x-2-2018}{2018}+\dfrac{x-1-2019}{2019}=0\)
\(\Leftrightarrow\dfrac{x-2020}{2017}+\dfrac{x-2020}{2018}+\dfrac{x-2020}{2019}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\right)=0\)
Vì \(\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\ne0\)
\(\Leftrightarrow x-2020=0\)
\(\Leftrightarrow x=2020\)
Vậy....
\(\frac{x-3}{2017}\)+\(\frac{x-2}{2018}\)+\(\frac{x-1}{2019}\)=3
= 4074342(x-3)+4072323(x-2)+4070306(x-1)=24653843442
=07342x- 12223026+ 4072323x-8144646+4070306x- 4070306= 24653843442
12216971x- 24437978= 24653843442
12216971x=24653843442+24437978
12216971x= 24678281420
x= 2020
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\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+2017}{3}+\frac{x+2016}{4}\)
\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+2017}{3}+1+\frac{x+2016}{4}+1\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)
Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy...
=-2016,00099
\(\frac{x}{2017}+\frac{x-1}{2017}=\frac{x-2}{2019}-1\)
\(\Leftrightarrow\frac{2x-1}{2017}=\frac{x-2021}{2019}\)
\(\Leftrightarrow4038x-2019=2017x-4076357\)
\(\Leftrightarrow2021x=4074338\)
\(\Leftrightarrow x=\frac{4074338}{2021}\)(nếu sai thì ib vs mik nha)