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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.
\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)
\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)
\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)
\(\Leftrightarrow\left(\sqrt{x-2016}-1\right)^2+\left(\sqrt{y-2017}-1\right)^2+\left(\sqrt{z-2018}-1\right)^2=0\)
\(ĐK:x\ge2016;y\ge2017;z\ge2018\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\y=2018\\z=2019\end{cases}}}\)
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
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
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+2017}{3}+\frac{x+2016}{4}\)
\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+2017}{3}+1+\frac{x+2016}{4}+1\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)
Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy...
\(\frac{2-x}{2016}-1=\frac{1-x}{2017}+\frac{x}{2018}\)
\(\Rightarrow\frac{2-x}{2016}-1=\frac{1-2018x}{4070306}+\frac{2017x}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}-1=\frac{1-2018x+2017x}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}-1=\frac{1-x}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}-1+1=\frac{1-x}{4070306}+1\)
\(\Rightarrow\frac{2-x}{2016}=\frac{1-x+4070306}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}=\frac{4070307-x}{4070306}\)
\(\Rightarrow4070306.\left(2-x\right)=2016.\left(4070307-x\right)\)
\(\Rightarrow8140612-4070306x=8205738912-2016x\)
\(\Rightarrow-4070306x+2016x=8205738912-8140612\)
\(\Rightarrow-4068290x=8197598300\)
\(\Rightarrow x=4,95\)
Vậy x=4,95
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