Cộng 2 vế của phương trình với 2 ta có: \(\frac{2-x}{2016}+1=\left(\frac{1-x}{2017}+1\right)-\left(\frac{x}{2018}-1\right)\)

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

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

Vì \(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)\(\Rightarrow2018-x=0\)\(\Leftrightarrow x=2018\)

Vậy tập nghiệm của phương trình là \(S=\left\{2018\right\}\)

\(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}\)

\(\Leftrightarrow\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}-\frac{x+2015}{5}-\frac{x+2016}{4}-\frac{x+2017}{3}=0\)

\(\Leftrightarrow\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)-\left(\frac{x+2015}{5}+1\right)-\left(\frac{x+2016}{4}+1\right)\)

\(-\left(\frac{x+2017}{3}+1\right)=0\)

\(\Leftrightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{5}-\frac{x+2020}{4}-\frac{x+2020}{3}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x+2020=0\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

<=> x=-2020

Vậy x=-2020

Giải phương trình: (3x-2)(x-1)^2(3x+8)=-16

\(25a^2-5a^3-a+5\)

\(=-5a^2\left(a-5\right)-\left(a-5\right)\)

\(=-\left(a-5\right)\left(5a^2+1\right)\)

((x - 2)(x + 10))/3 - ((x + 40)(x + 10))/12 = ((x - 2)(x + 4))/4

<=> 4(x - 2)(x + 10) - (x + 4)(x + 10) = 3(x - 2)(x + 4)

<=> 4x^2 + 40x - 8x - 80 - x^2 - 10x - 4x - 40 = 3x^2 + 12x - 6x - 24

<=> 3x^2 + 18x - 120 = 3x^2 + 6x - 24

<=> 18x - 120 = 6x - 24

<=> 18x - 120 - 6x = -24

<=> 12x - 120 = -24

<=> 12x = -24 + 120

<=> 12x = 96

<=> x = 8

\(\left(x^2+3x+2\right)\left(x^2+5x+6\right)=\left(x^2+2x+x+2\right)\left(x^2+2x+3x+6\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x+2\right)\left(x+3\right)=\left[\left(x+1\right)\left(x+3\right)\right]\left(x+2\right)^2\)

\(=\left(x^2+4x+3\right)\left(x^2+4x+4\right).\text{Đặt: }x^2+4x+3\Rightarrow a\left(a+1\right)=72\)

\(\text{cái này bạn giải ra được:}a=8\text{ hoặc }a=-9\text{ thấy:}a+1=\left(x+2\right)^2\ge0\Rightarrow a\ge-1\Rightarrow a=8\)

\(\Leftrightarrow\left(x+2\right)^2=9\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)