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\(\frac{x}{2017}+\frac{x-1}{2017}=\frac{x-2}{2019}-1\)
\(\Leftrightarrow\frac{2x-1}{2017}=\frac{x-2021}{2019}\)
\(\Leftrightarrow4038x-2019=2017x-4076357\)
\(\Leftrightarrow2021x=4074338\)
\(\Leftrightarrow x=\frac{4074338}{2021}\)(nếu sai thì ib vs mik nha)
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\(\Leftrightarrow x\left(x-2017\right)-\left(x-2017\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2017\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2017=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2017\end{cases}}}\)
Vậy.....
\(x\left(x-2017\right)-x+2017=0\)
\(\Leftrightarrow x\left(x-2017\right)-\left(x-2017\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2017\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2017=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2017\end{cases}}\)
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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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Ta co:
\(\left|x-2016\right|+\left|x-2018\right|=\left|x-2016\right|+\left|2018-x\right|\ge\left|x-2016+2018-x\right|=2\)
\(\left|x-2017\right|\ge0\)
\(\Rightarrow\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\ge2\)
Dau "=" xay ra tai \(\hept{\begin{cases}2016\le x\le2018\\x=2017\end{cases}}\)
Vay x=2017
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\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
<=> \(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
<=> \(\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
<=> x + 2015 = 0 ( vì \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x = - 2015
Vậy x = -2015.
Giải phương trình :
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
\(\Rightarrow\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Mà \(\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)>0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)