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.

Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)

\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)

Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên 2x + 4 = 0

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

Vậy, x = -2

\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)

\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)

Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên \(2x+4=0\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

Vậy, x = -2

a) \(22-x\left(1-4x\right)=\left(2x+3\right)^3\)

\(\Leftrightarrow22-x+4x^2=8x^3+36x^2+54x+27\)

\(\Leftrightarrow-x-54x+4x^2-36x^2-8x^3=-22+27\)

\(\Leftrightarrow-8x^3-32x^2-55x=5\Leftrightarrow-8x^3-32x^2-55x-5=0\)

Bn tự làm tiếp nhé

b) \(\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\Leftrightarrow\frac{2.2x}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\)

\(\Leftrightarrow2.2x+2x-1=2\left(4-x\right)\Leftrightarrow4x+2x-1=8-2x\)

\(\Leftrightarrow6x-1=8-2x\Leftrightarrow8x=9\Leftrightarrow x=\frac{9}{8}\)

Vậy phương trình có tập nghiệm S ={9/8}

c) \(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)

\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)

\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)

\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Do \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}>0\)

Nên \(x-2020=0\Leftrightarrow x=2020\)

2x²+y²+9=6x+2xy

=>2x²+y²+9-6x-2xy=0

=>(x²-2xy+y²)+(x²-6x+9)=0

=>(x-y)²+(x-3)²=0

do (x-y)² ≥ 0 ∀ x,y

(x-3)² ≥ 0 ∀x

=>(x-y)²+(x-3)² =0 khi

=>\(\left[{}\begin{matrix}x-y=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=x=3\\x=3\end{matrix}\right.\)

thay x=3 và y=3

Q=3²⁰¹⁷.3²⁰¹⁸-3²⁰¹⁸. 3²⁰¹⁷+\(\dfrac{1}{9}.3.3\)

Q=1

bạn giỏi quá!

d, 2x²-5x-3 = 0

\(\Leftrightarrow\)2x²-6x+x-3= 0

\(\Leftrightarrow\)(2x²-6x) +(x-3) = 0

\(\Leftrightarrow\)2x(x-3) + (x-3) = 0

\(\Leftrightarrow\)(x-3) (2x+1) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S =\(\left\{3;\frac{-1}{2}\right\}\)

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

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

\(\Leftrightarrow11x-6049=0\)

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

Sai r bạn ơi = -2015/11

\(5x^2+5y^2+8xy-2x+2y+2=0\Leftrightarrow x^2+4x^2+y^2+4y^2+8xy-2x+2y+1+1=0\Leftrightarrow\left(x^2-2x+1\right)+\left(4x^2+8xy+4y^2\right)+\left(y^2+2y+1\right)=0\Leftrightarrow\left(x-1\right)^2+4\left(x+y\right)^2+\left(y+1\right)^2=0\)

Mà \(\left\{{}4\begin{matrix}\left(x-1\right)^2\ge0\\\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\4\left(x+y\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+y=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-y\\y=-1\end{matrix}\right.\)

Với \(x=1;y=-1\) ta có:

\(M=\left(x+y\right)^{2016}+\left(x-2\right)^{2017}+\left(y+1\right)^{2018}=\left(1-1\right)^{2016}+\left(1-2\right)^{2017}+\left(-1+1\right)^{2018}=0+\left(-1\right)+0=-1\)

Vậy M = -1