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.

\(\left|x-2015\right|^{2016}+\left|x-2016\right|^{2017}=1\)

Có: \(\left|x-2015\right|^{2016}\ge0;\left|x-2016\right|^{2017}\ge0\)

TH1: \(\hept{\begin{cases}\left|x-2015\right|^{2016}=1\\\left|x-2016\right|^{2017}=0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x-2015\right|=1\\\left|x-2016\right|=0\end{cases}}\)

THa: \(x-2015=-1\Rightarrow x=2014\)

Thay vào: \(2014-2016\ne0\) ( loại)

THb: \(x-2015=1\Rightarrow x=2016\)

Thay vào:  \(2016-2016=0\)( chọn )

TH2: \(\hept{\begin{cases}\left|x-2015\right|^{2016}=0\\\left|x-2016\right|^{2017}=1\end{cases}}\Rightarrow\hept{\begin{cases}\left|x-2015\right|=0\\\left|x-2016\right|=1\end{cases}}\)

THc: \(x-2016=-1\Rightarrow x=2015\)

Thay vào:  \(2015-2015=0\)( chọn )

THd: \(x-2016=1\Rightarrow x=2017\)

Thay vào: \(2017-2015\ne0\)

Vậy: x = 2016 hoặc x = 2015

x=2015

\(\left|x-2015\right|^{2016}\ge0\)

\(\left|x-2016\right|^{2017}\ge0\)

\(\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}\)

Đặt 2x²+x-2015=a; x²-5x-2016=b

phương trình tương đương a²+4b²=4ab

=> a²-4ab+4b²=0

=> (a-2b)²=0

=> a=2b

vậy 2x²+x-2015=2*(x²-5x-2016)

=> x=\(\frac{-2017}{11}\)

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

\(\Leftrightarrow\left(2x^2+x-2017\right)^2-4\left(2x^2+x-2017\right)\left(x^2-5x-2016\right)+4\left(x^2-5x-2016\right)^2=0\)

\(\Leftrightarrow\left(2x^2+x-2017-2\left(x^2-5x-2016\right)\right)^2=0\)

\(\Leftrightarrow11x-6049=0\)

\(\Rightarrow x=\frac{6049}{11}\)

Xét: \(\sqrt{1+n^2+\frac{n^2}{\left(n+1\right)^2}}=\sqrt{\frac{\left(n+1\right)^2+n^2\left(n+1\right)^2+n^2}{\left(n+1\right)^2}}\) (với \(n\inℕ\))

\(=\sqrt{\frac{n^2+2n+1+n^4+2n^3+n^2+n^2}{\left(n+1\right)^2}}\)

\(=\sqrt{\frac{n^4+n^2+1+2n^3+2n^2+2n}{\left(n+1\right)^2}}\)

\(=\sqrt{\frac{\left(n^2+n+1\right)^2}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}\)

Áp dụng vào ta tính được: \(\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}=2015+\frac{1}{2016}+\frac{2015}{2016}\)

\(=2015+1=2016\)

Khi đó: \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=2016\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2016\)

Đến đây xét tiếp các TH nhé, ez rồi:))

chẳng biết đúng ko,mới lớp 5

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\sqrt{x^2}-\sqrt{2x}+\sqrt{1}+\sqrt{x^2}-\sqrt{4x}+\sqrt{4}=\sqrt{1}+\sqrt{2015^2}+\sqrt{\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\sqrt{x^2}-\sqrt{6x}+3=1+2015+\frac{2015}{2016}+\frac{2015}{2016}\)

\(x-\sqrt{6x}=1+\frac{2015}{1+2016+2016}-3\)

\(x-\sqrt{6x}=2-\frac{2015}{4033}\)

\(x-\sqrt{6x}=\frac{6051}{4033}\)

pt <=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=\sqrt{1+2.2015+2015^2-2.2015+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\Leftrightarrow\left(x-1\right)+\left(x-2\right)=\sqrt{2016^2-2.2016.\frac{2015}{2016}+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=\sqrt{\left(2016-\frac{2015}{2016}\right)^2}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=2016-\frac{2015}{2016}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=2016\)

\(\Leftrightarrow2x=2019\)

\(\Leftrightarrow x=\frac{2019}{2}\)